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---
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comments: true
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---
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# 7.4. 二叉搜索树
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「二叉搜索树 Binary Search Tree」满足以下条件:
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1. 对于根节点,左子树中所有节点的值 $<$ 根节点的值 $<$ 右子树中所有节点的值;
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2. 任意节点的左、右子树也是二叉搜索树,即同样满足条件 `1.` ;
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![二叉搜索树](binary_search_tree.assets/binary_search_tree.png)
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<p align="center"> Fig. 二叉搜索树 </p>
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## 7.4.1. 二叉搜索树的操作
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### 查找节点
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给定目标节点值 `num` ,可以根据二叉搜索树的性质来查找。我们声明一个节点 `cur` ,从二叉树的根节点 `root` 出发,循环比较节点值 `cur.val` 和 `num` 之间的大小关系
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- 若 `cur.val < num` ,说明目标节点在 `cur` 的右子树中,因此执行 `cur = cur.right` ;
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- 若 `cur.val > num` ,说明目标节点在 `cur` 的左子树中,因此执行 `cur = cur.left` ;
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- 若 `cur.val = num` ,说明找到目标节点,跳出循环并返回该节点;
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=== "<1>"
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![二叉搜索树查找节点示例](binary_search_tree.assets/bst_search_step1.png)
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=== "<2>"
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![bst_search_step2](binary_search_tree.assets/bst_search_step2.png)
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=== "<3>"
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![bst_search_step3](binary_search_tree.assets/bst_search_step3.png)
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=== "<4>"
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![bst_search_step4](binary_search_tree.assets/bst_search_step4.png)
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二叉搜索树的查找操作与二分查找算法的工作原理一致,都是每轮排除一半情况。循环次数最多为二叉树的高度,当二叉树平衡时,使用 $O(\log n)$ 时间。
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=== "Java"
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```java title="binary_search_tree.java"
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/* 查找节点 */
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TreeNode search(int num) {
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TreeNode cur = root;
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// 循环查找,越过叶节点后跳出
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while (cur != null) {
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// 目标节点在 cur 的右子树中
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if (cur.val < num)
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cur = cur.right;
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// 目标节点在 cur 的左子树中
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else if (cur.val > num)
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cur = cur.left;
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// 找到目标节点,跳出循环
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else
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break;
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}
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// 返回目标节点
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return cur;
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}
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```
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=== "C++"
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```cpp title="binary_search_tree.cpp"
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/* 查找节点 */
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TreeNode *search(int num) {
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TreeNode *cur = root;
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// 循环查找,越过叶节点后跳出
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while (cur != nullptr) {
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// 目标节点在 cur 的右子树中
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if (cur->val < num)
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cur = cur->right;
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// 目标节点在 cur 的左子树中
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else if (cur->val > num)
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cur = cur->left;
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// 找到目标节点,跳出循环
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else
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break;
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}
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// 返回目标节点
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return cur;
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}
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```
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=== "Python"
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```python title="binary_search_tree.py"
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def search(self, num: int) -> TreeNode | None:
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"""查找节点"""
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cur: TreeNode | None = self.root
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# 循环查找,越过叶节点后跳出
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while cur is not None:
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# 目标节点在 cur 的右子树中
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if cur.val < num:
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cur = cur.right
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# 目标节点在 cur 的左子树中
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elif cur.val > num:
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cur = cur.left
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# 找到目标节点,跳出循环
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else:
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break
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return cur
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```
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=== "Go"
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```go title="binary_search_tree.go"
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/* 查找节点 */
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func (bst *binarySearchTree) search(num int) *TreeNode {
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node := bst.root
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// 循环查找,越过叶节点后跳出
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for node != nil {
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if node.Val < num {
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// 目标节点在 cur 的右子树中
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node = node.Right
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} else if node.Val > num {
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// 目标节点在 cur 的左子树中
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node = node.Left
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} else {
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// 找到目标节点,跳出循环
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break
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}
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}
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// 返回目标节点
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return node
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}
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```
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=== "JavaScript"
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```javascript title="binary_search_tree.js"
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/* 查找节点 */
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function search(num) {
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let cur = root;
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// 循环查找,越过叶节点后跳出
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while (cur !== null) {
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// 目标节点在 cur 的右子树中
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if (cur.val < num) cur = cur.right;
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// 目标节点在 cur 的左子树中
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else if (cur.val > num) cur = cur.left;
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// 找到目标节点,跳出循环
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else break;
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}
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// 返回目标节点
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return cur;
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}
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```
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=== "TypeScript"
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```typescript title="binary_search_tree.ts"
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/* 查找节点 */
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function search(num: number): TreeNode | null {
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let cur = root;
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// 循环查找,越过叶节点后跳出
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while (cur !== null) {
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if (cur.val < num) {
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cur = cur.right; // 目标节点在 cur 的右子树中
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} else if (cur.val > num) {
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cur = cur.left; // 目标节点在 cur 的左子树中
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} else {
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break; // 找到目标节点,跳出循环
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}
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}
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// 返回目标节点
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return cur;
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}
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```
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=== "C"
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```c title="binary_search_tree.c"
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/* 查找节点 */
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TreeNode *search(binarySearchTree *bst, int num) {
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TreeNode *cur = bst->root;
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// 循环查找,越过叶节点后跳出
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while (cur != NULL) {
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if (cur->val < num) {
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// 目标节点在 cur 的右子树中
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cur = cur->right;
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} else if (cur->val > num) {
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// 目标节点在 cur 的左子树中
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cur = cur->left;
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} else {
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// 找到目标节点,跳出循环
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break;
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}
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}
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// 返回目标节点
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return cur;
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}
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```
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=== "C#"
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```csharp title="binary_search_tree.cs"
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/* 查找节点 */
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TreeNode? search(int num) {
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TreeNode? cur = root;
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// 循环查找,越过叶节点后跳出
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while (cur != null) {
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// 目标节点在 cur 的右子树中
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if (cur.val < num) cur =
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cur.right;
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// 目标节点在 cur 的左子树中
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else if (cur.val > num)
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cur = cur.left;
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// 找到目标节点,跳出循环
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else
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break;
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}
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// 返回目标节点
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return cur;
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}
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```
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=== "Swift"
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```swift title="binary_search_tree.swift"
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/* 查找节点 */
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func search(num: Int) -> TreeNode? {
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var cur = root
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// 循环查找,越过叶节点后跳出
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while cur != nil {
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// 目标节点在 cur 的右子树中
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if cur!.val < num {
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cur = cur?.right
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}
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// 目标节点在 cur 的左子树中
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else if cur!.val > num {
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cur = cur?.left
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}
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// 找到目标节点,跳出循环
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else {
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break
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}
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}
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// 返回目标节点
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return cur
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}
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```
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=== "Zig"
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```zig title="binary_search_tree.zig"
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// 查找节点
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fn search(self: *Self, num: T) ?*inc.TreeNode(T) {
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var cur = self.root;
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// 循环查找,越过叶节点后跳出
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while (cur != null) {
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// 目标节点在 cur 的右子树中
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if (cur.?.val < num) {
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cur = cur.?.right;
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// 目标节点在 cur 的左子树中
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} else if (cur.?.val > num) {
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cur = cur.?.left;
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// 找到目标节点,跳出循环
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} else {
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break;
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}
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}
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// 返回目标节点
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return cur;
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}
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```
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=== "Dart"
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```dart title="binary_search_tree.dart"
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[class]{BinarySearchTree}-[func]{search}
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```
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### 插入节点
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给定一个待插入元素 `num` ,为了保持二叉搜索树“左子树 < 根节点 < 右子树”的性质,插入操作分为两步:
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1. **查找插入位置**:与查找操作相似,从根节点出发,根据当前节点值和 `num` 的大小关系循环向下搜索,直到越过叶节点(遍历至 $\text{None}$ )时跳出循环;
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2. **在该位置插入节点**:初始化节点 `num` ,将该节点置于 $\text{None}$ 的位置;
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二叉搜索树不允许存在重复节点,否则将违反其定义。因此,若待插入节点在树中已存在,则不执行插入,直接返回。
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![在二叉搜索树中插入节点](binary_search_tree.assets/bst_insert.png)
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<p align="center"> Fig. 在二叉搜索树中插入节点 </p>
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=== "Java"
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```java title="binary_search_tree.java"
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/* 插入节点 */
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void insert(int num) {
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// 若树为空,直接提前返回
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if (root == null)
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return;
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TreeNode cur = root, pre = null;
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// 循环查找,越过叶节点后跳出
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while (cur != null) {
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// 找到重复节点,直接返回
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if (cur.val == num)
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return;
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pre = cur;
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// 插入位置在 cur 的右子树中
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if (cur.val < num)
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cur = cur.right;
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// 插入位置在 cur 的左子树中
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else
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cur = cur.left;
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}
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// 插入节点
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TreeNode node = new TreeNode(num);
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if (pre.val < num)
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pre.right = node;
|
|
|
|
|
else
|
|
|
|
|
pre.left = node;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C++"
|
|
|
|
|
|
|
|
|
|
```cpp title="binary_search_tree.cpp"
|
|
|
|
|
/* 插入节点 */
|
|
|
|
|
void insert(int num) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (root == nullptr)
|
|
|
|
|
return;
|
|
|
|
|
TreeNode *cur = root, *pre = nullptr;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur != nullptr) {
|
|
|
|
|
// 找到重复节点,直接返回
|
|
|
|
|
if (cur->val == num)
|
|
|
|
|
return;
|
|
|
|
|
pre = cur;
|
|
|
|
|
// 插入位置在 cur 的右子树中
|
|
|
|
|
if (cur->val < num)
|
|
|
|
|
cur = cur->right;
|
|
|
|
|
// 插入位置在 cur 的左子树中
|
|
|
|
|
else
|
|
|
|
|
cur = cur->left;
|
|
|
|
|
}
|
|
|
|
|
// 插入节点
|
|
|
|
|
TreeNode *node = new TreeNode(num);
|
|
|
|
|
if (pre->val < num)
|
|
|
|
|
pre->right = node;
|
|
|
|
|
else
|
|
|
|
|
pre->left = node;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="binary_search_tree.py"
|
|
|
|
|
def insert(self, num: int) -> None:
|
|
|
|
|
"""插入节点"""
|
|
|
|
|
# 若树为空,直接提前返回
|
|
|
|
|
if self.root is None:
|
|
|
|
|
return
|
|
|
|
|
|
|
|
|
|
# 循环查找,越过叶节点后跳出
|
|
|
|
|
cur, pre = self.root, None
|
|
|
|
|
while cur is not None:
|
|
|
|
|
# 找到重复节点,直接返回
|
|
|
|
|
if cur.val == num:
|
|
|
|
|
return
|
|
|
|
|
pre = cur
|
|
|
|
|
# 插入位置在 cur 的右子树中
|
|
|
|
|
if cur.val < num:
|
|
|
|
|
cur = cur.right
|
|
|
|
|
# 插入位置在 cur 的左子树中
|
|
|
|
|
else:
|
|
|
|
|
cur = cur.left
|
|
|
|
|
|
|
|
|
|
# 插入节点
|
|
|
|
|
node = TreeNode(num)
|
|
|
|
|
if pre.val < num:
|
|
|
|
|
pre.right = node
|
|
|
|
|
else:
|
|
|
|
|
pre.left = node
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="binary_search_tree.go"
|
|
|
|
|
/* 插入节点 */
|
|
|
|
|
func (bst *binarySearchTree) insert(num int) {
|
|
|
|
|
cur := bst.root
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if cur == nil {
|
|
|
|
|
return
|
|
|
|
|
}
|
|
|
|
|
// 待插入节点之前的节点位置
|
|
|
|
|
var pre *TreeNode = nil
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
for cur != nil {
|
|
|
|
|
if cur.Val == num {
|
|
|
|
|
return
|
|
|
|
|
}
|
|
|
|
|
pre = cur
|
|
|
|
|
if cur.Val < num {
|
|
|
|
|
cur = cur.Right
|
|
|
|
|
} else {
|
|
|
|
|
cur = cur.Left
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 插入节点
|
|
|
|
|
node := NewTreeNode(num)
|
|
|
|
|
if pre.Val < num {
|
|
|
|
|
pre.Right = node
|
|
|
|
|
} else {
|
|
|
|
|
pre.Left = node
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```javascript title="binary_search_tree.js"
|
|
|
|
|
/* 插入节点 */
|
|
|
|
|
function insert(num) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (root === null) return;
|
|
|
|
|
let cur = root,
|
|
|
|
|
pre = null;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur !== null) {
|
|
|
|
|
// 找到重复节点,直接返回
|
|
|
|
|
if (cur.val === num) return;
|
|
|
|
|
pre = cur;
|
|
|
|
|
// 插入位置在 cur 的右子树中
|
|
|
|
|
if (cur.val < num) cur = cur.right;
|
|
|
|
|
// 插入位置在 cur 的左子树中
|
|
|
|
|
else cur = cur.left;
|
|
|
|
|
}
|
|
|
|
|
// 插入节点
|
|
|
|
|
let node = new TreeNode(num);
|
|
|
|
|
if (pre.val < num) pre.right = node;
|
|
|
|
|
else pre.left = node;
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="binary_search_tree.ts"
|
|
|
|
|
/* 插入节点 */
|
|
|
|
|
function insert(num: number): void {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (root === null) {
|
|
|
|
|
return;
|
|
|
|
|
}
|
|
|
|
|
let cur = root,
|
|
|
|
|
pre: TreeNode | null = null;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur !== null) {
|
|
|
|
|
if (cur.val === num) {
|
|
|
|
|
return; // 找到重复节点,直接返回
|
|
|
|
|
}
|
|
|
|
|
pre = cur;
|
|
|
|
|
if (cur.val < num) {
|
|
|
|
|
cur = cur.right as TreeNode; // 插入位置在 cur 的右子树中
|
|
|
|
|
} else {
|
|
|
|
|
cur = cur.left as TreeNode; // 插入位置在 cur 的左子树中
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 插入节点
|
|
|
|
|
let node = new TreeNode(num);
|
|
|
|
|
if (pre!.val < num) {
|
|
|
|
|
pre!.right = node;
|
|
|
|
|
} else {
|
|
|
|
|
pre!.left = node;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="binary_search_tree.c"
|
|
|
|
|
/* 插入节点 */
|
|
|
|
|
void insert(binarySearchTree *bst, int num) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (bst->root == NULL)
|
|
|
|
|
return;
|
|
|
|
|
TreeNode *cur = bst->root, *pre = NULL;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur != NULL) {
|
|
|
|
|
// 找到重复节点,直接返回
|
|
|
|
|
if (cur->val == num) {
|
|
|
|
|
return;
|
|
|
|
|
}
|
|
|
|
|
pre = cur;
|
|
|
|
|
if (cur->val < num) {
|
|
|
|
|
// 插入位置在 cur 的右子树中
|
|
|
|
|
cur = cur->right;
|
|
|
|
|
} else {
|
|
|
|
|
// 插入位置在 cur 的左子树中
|
|
|
|
|
cur = cur->left;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 插入节点
|
|
|
|
|
TreeNode *node = newTreeNode(num);
|
|
|
|
|
if (pre->val < num) {
|
|
|
|
|
pre->right = node;
|
|
|
|
|
} else {
|
|
|
|
|
pre->left = node;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="binary_search_tree.cs"
|
|
|
|
|
/* 插入节点 */
|
|
|
|
|
void insert(int num) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (root == null)
|
|
|
|
|
return;
|
|
|
|
|
TreeNode? cur = root, pre = null;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur != null) {
|
|
|
|
|
// 找到重复节点,直接返回
|
|
|
|
|
if (cur.val == num)
|
|
|
|
|
return;
|
|
|
|
|
pre = cur;
|
|
|
|
|
// 插入位置在 cur 的右子树中
|
|
|
|
|
if (cur.val < num)
|
|
|
|
|
cur = cur.right;
|
|
|
|
|
// 插入位置在 cur 的左子树中
|
|
|
|
|
else
|
|
|
|
|
cur = cur.left;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
// 插入节点
|
|
|
|
|
TreeNode node = new TreeNode(num);
|
|
|
|
|
if (pre != null) {
|
|
|
|
|
if (pre.val < num)
|
|
|
|
|
pre.right = node;
|
|
|
|
|
else
|
|
|
|
|
pre.left = node;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="binary_search_tree.swift"
|
|
|
|
|
/* 插入节点 */
|
|
|
|
|
func insert(num: Int) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if root == nil {
|
|
|
|
|
return
|
|
|
|
|
}
|
|
|
|
|
var cur = root
|
|
|
|
|
var pre: TreeNode?
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while cur != nil {
|
|
|
|
|
// 找到重复节点,直接返回
|
|
|
|
|
if cur!.val == num {
|
|
|
|
|
return
|
|
|
|
|
}
|
|
|
|
|
pre = cur
|
|
|
|
|
// 插入位置在 cur 的右子树中
|
|
|
|
|
if cur!.val < num {
|
|
|
|
|
cur = cur?.right
|
|
|
|
|
}
|
|
|
|
|
// 插入位置在 cur 的左子树中
|
|
|
|
|
else {
|
|
|
|
|
cur = cur?.left
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 插入节点
|
|
|
|
|
let node = TreeNode(x: num)
|
|
|
|
|
if pre!.val < num {
|
|
|
|
|
pre?.right = node
|
|
|
|
|
} else {
|
|
|
|
|
pre?.left = node
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="binary_search_tree.zig"
|
|
|
|
|
// 插入节点
|
|
|
|
|
fn insert(self: *Self, num: T) !void {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (self.root == null) return;
|
|
|
|
|
var cur = self.root;
|
|
|
|
|
var pre: ?*inc.TreeNode(T) = null;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur != null) {
|
|
|
|
|
// 找到重复节点,直接返回
|
|
|
|
|
if (cur.?.val == num) return;
|
|
|
|
|
pre = cur;
|
|
|
|
|
// 插入位置在 cur 的右子树中
|
|
|
|
|
if (cur.?.val < num) {
|
|
|
|
|
cur = cur.?.right;
|
|
|
|
|
// 插入位置在 cur 的左子树中
|
|
|
|
|
} else {
|
|
|
|
|
cur = cur.?.left;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 插入节点
|
|
|
|
|
var node = try self.mem_allocator.create(inc.TreeNode(T));
|
|
|
|
|
node.init(num);
|
|
|
|
|
if (pre.?.val < num) {
|
|
|
|
|
pre.?.right = node;
|
|
|
|
|
} else {
|
|
|
|
|
pre.?.left = node;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="binary_search_tree.dart"
|
|
|
|
|
[class]{BinarySearchTree}-[func]{insert}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
为了插入节点,我们需要利用辅助节点 `pre` 保存上一轮循环的节点,这样在遍历至 $\text{None}$ 时,我们可以获取到其父节点,从而完成节点插入操作。
|
|
|
|
|
|
|
|
|
|
与查找节点相同,插入节点使用 $O(\log n)$ 时间。
|
|
|
|
|
|
|
|
|
|
### 删除节点
|
|
|
|
|
|
|
|
|
|
与插入节点类似,我们需要在删除操作后维持二叉搜索树的“左子树 < 根节点 < 右子树”的性质。首先,我们需要在二叉树中执行查找操作,获取待删除节点。接下来,根据待删除节点的子节点数量,删除操作需分为三种情况:
|
|
|
|
|
|
|
|
|
|
当待删除节点的度为 $0$ 时,表示待删除节点是叶节点,可以直接删除。
|
|
|
|
|
|
|
|
|
|
![在二叉搜索树中删除节点(度为 0)](binary_search_tree.assets/bst_remove_case1.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 在二叉搜索树中删除节点(度为 0) </p>
|
|
|
|
|
|
|
|
|
|
当待删除节点的度为 $1$ 时,将待删除节点替换为其子节点即可。
|
|
|
|
|
|
|
|
|
|
![在二叉搜索树中删除节点(度为 1)](binary_search_tree.assets/bst_remove_case2.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 在二叉搜索树中删除节点(度为 1) </p>
|
|
|
|
|
|
|
|
|
|
当待删除节点的度为 $2$ 时,我们无法直接删除它,而需要使用一个节点替换该节点。由于要保持二叉搜索树“左 $<$ 根 $<$ 右”的性质,因此这个节点可以是右子树的最小节点或左子树的最大节点。假设我们选择右子树的最小节点(或者称为中序遍历的下个节点),则删除操作为:
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1. 找到待删除节点在“中序遍历序列”中的下一个节点,记为 `tmp` ;
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2. 将 `tmp` 的值覆盖待删除节点的值,并在树中递归删除节点 `tmp` ;
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=== "<1>"
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![二叉搜索树删除节点示例](binary_search_tree.assets/bst_remove_case3_step1.png)
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=== "<2>"
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![bst_remove_case3_step2](binary_search_tree.assets/bst_remove_case3_step2.png)
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=== "<3>"
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![bst_remove_case3_step3](binary_search_tree.assets/bst_remove_case3_step3.png)
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=== "<4>"
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![bst_remove_case3_step4](binary_search_tree.assets/bst_remove_case3_step4.png)
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删除节点操作同样使用 $O(\log n)$ 时间,其中查找待删除节点需要 $O(\log n)$ 时间,获取中序遍历后继节点需要 $O(\log n)$ 时间。
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=== "Java"
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```java title="binary_search_tree.java"
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/* 删除节点 */
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void remove(int num) {
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// 若树为空,直接提前返回
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if (root == null)
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return;
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TreeNode cur = root, pre = null;
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// 循环查找,越过叶节点后跳出
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while (cur != null) {
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// 找到待删除节点,跳出循环
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if (cur.val == num)
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break;
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pre = cur;
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// 待删除节点在 cur 的右子树中
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if (cur.val < num)
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cur = cur.right;
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// 待删除节点在 cur 的左子树中
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else
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cur = cur.left;
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}
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// 若无待删除节点,则直接返回
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if (cur == null)
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return;
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// 子节点数量 = 0 or 1
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if (cur.left == null || cur.right == null) {
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// 当子节点数量 = 0 / 1 时, child = null / 该子节点
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TreeNode child = cur.left != null ? cur.left : cur.right;
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// 删除节点 cur
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if (cur != root) {
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if (pre.left == cur)
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pre.left = child;
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else
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pre.right = child;
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} else {
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// 若删除节点为根节点,则重新指定根节点
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root = child;
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}
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}
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// 子节点数量 = 2
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else {
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// 获取中序遍历中 cur 的下一个节点
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TreeNode tmp = cur.right;
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while (tmp.left != null) {
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tmp = tmp.left;
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}
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// 递归删除节点 tmp
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remove(tmp.val);
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// 用 tmp 覆盖 cur
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cur.val = tmp.val;
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}
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}
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```
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=== "C++"
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```cpp title="binary_search_tree.cpp"
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|
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/* 删除节点 */
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void remove(int num) {
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// 若树为空,直接提前返回
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if (root == nullptr)
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return;
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TreeNode *cur = root, *pre = nullptr;
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// 循环查找,越过叶节点后跳出
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while (cur != nullptr) {
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// 找到待删除节点,跳出循环
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if (cur->val == num)
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break;
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pre = cur;
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// 待删除节点在 cur 的右子树中
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if (cur->val < num)
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cur = cur->right;
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// 待删除节点在 cur 的左子树中
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else
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cur = cur->left;
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}
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|
|
// 若无待删除节点,则直接返回
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|
|
if (cur == nullptr)
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|
return;
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// 子节点数量 = 0 or 1
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|
|
if (cur->left == nullptr || cur->right == nullptr) {
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// 当子节点数量 = 0 / 1 时, child = nullptr / 该子节点
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|
|
TreeNode *child = cur->left != nullptr ? cur->left : cur->right;
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|
// 删除节点 cur
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|
|
if (cur != root) {
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|
|
if (pre->left == cur)
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pre->left = child;
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else
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pre->right = child;
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} else {
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// 若删除节点为根节点,则重新指定根节点
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root = child;
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}
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|
// 释放内存
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|
delete cur;
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}
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|
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// 子节点数量 = 2
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else {
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// 获取中序遍历中 cur 的下一个节点
|
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|
|
TreeNode *tmp = cur->right;
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|
|
while (tmp->left != nullptr) {
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|
|
tmp = tmp->left;
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|
|
}
|
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|
|
int tmpVal = tmp->val;
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|
|
// 递归删除节点 tmp
|
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|
|
remove(tmp->val);
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|
|
// 用 tmp 覆盖 cur
|
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|
|
cur->val = tmpVal;
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|
|
}
|
|
|
|
|
}
|
|
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|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Python"
|
|
|
|
|
|
|
|
|
|
```python title="binary_search_tree.py"
|
|
|
|
|
def remove(self, num: int) -> None:
|
|
|
|
|
"""删除节点"""
|
|
|
|
|
# 若树为空,直接提前返回
|
|
|
|
|
if self.root is None:
|
|
|
|
|
return
|
|
|
|
|
|
|
|
|
|
# 循环查找,越过叶节点后跳出
|
|
|
|
|
cur, pre = self.root, None
|
|
|
|
|
while cur is not None:
|
|
|
|
|
# 找到待删除节点,跳出循环
|
|
|
|
|
if cur.val == num:
|
|
|
|
|
break
|
|
|
|
|
pre = cur
|
|
|
|
|
# 待删除节点在 cur 的右子树中
|
|
|
|
|
if cur.val < num:
|
|
|
|
|
cur = cur.right
|
|
|
|
|
# 待删除节点在 cur 的左子树中
|
|
|
|
|
else:
|
|
|
|
|
cur = cur.left
|
|
|
|
|
# 若无待删除节点,则直接返回
|
|
|
|
|
if cur is None:
|
|
|
|
|
return
|
|
|
|
|
|
|
|
|
|
# 子节点数量 = 0 or 1
|
|
|
|
|
if cur.left is None or cur.right is None:
|
|
|
|
|
# 当子节点数量 = 0 / 1 时, child = null / 该子节点
|
|
|
|
|
child = cur.left or cur.right
|
|
|
|
|
# 删除节点 cur
|
|
|
|
|
if cur != self.root:
|
|
|
|
|
if pre.left == cur:
|
|
|
|
|
pre.left = child
|
|
|
|
|
else:
|
|
|
|
|
pre.right = child
|
|
|
|
|
else:
|
|
|
|
|
# 若删除节点为根节点,则重新指定根节点
|
|
|
|
|
self.__root = cur
|
|
|
|
|
# 子节点数量 = 2
|
|
|
|
|
else:
|
|
|
|
|
# 获取中序遍历中 cur 的下一个节点
|
|
|
|
|
tmp: TreeNode = cur.right
|
|
|
|
|
while tmp.left is not None:
|
|
|
|
|
tmp = tmp.left
|
|
|
|
|
# 递归删除节点 tmp
|
|
|
|
|
self.remove(tmp.val)
|
|
|
|
|
# 用 tmp 覆盖 cur
|
|
|
|
|
cur.val = tmp.val
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Go"
|
|
|
|
|
|
|
|
|
|
```go title="binary_search_tree.go"
|
|
|
|
|
/* 删除节点 */
|
|
|
|
|
func (bst *binarySearchTree) remove(num int) {
|
|
|
|
|
cur := bst.root
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if cur == nil {
|
|
|
|
|
return
|
|
|
|
|
}
|
|
|
|
|
// 待删除节点之前的节点位置
|
|
|
|
|
var pre *TreeNode = nil
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
for cur != nil {
|
|
|
|
|
if cur.Val == num {
|
|
|
|
|
break
|
|
|
|
|
}
|
|
|
|
|
pre = cur
|
|
|
|
|
if cur.Val < num {
|
|
|
|
|
// 待删除节点在右子树中
|
|
|
|
|
cur = cur.Right
|
|
|
|
|
} else {
|
|
|
|
|
// 待删除节点在左子树中
|
|
|
|
|
cur = cur.Left
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 若无待删除节点,则直接返回
|
|
|
|
|
if cur == nil {
|
|
|
|
|
return
|
|
|
|
|
}
|
|
|
|
|
// 子节点数为 0 或 1
|
|
|
|
|
if cur.Left == nil || cur.Right == nil {
|
|
|
|
|
var child *TreeNode = nil
|
|
|
|
|
// 取出待删除节点的子节点
|
|
|
|
|
if cur.Left != nil {
|
|
|
|
|
child = cur.Left
|
|
|
|
|
} else {
|
|
|
|
|
child = cur.Right
|
|
|
|
|
}
|
|
|
|
|
// 删除节点 cur
|
|
|
|
|
if cur != bst.root {
|
|
|
|
|
if pre.Left == cur {
|
|
|
|
|
pre.Left = child
|
|
|
|
|
} else {
|
|
|
|
|
pre.Right = child
|
|
|
|
|
}
|
|
|
|
|
} else {
|
|
|
|
|
// 若删除节点为根节点,则重新指定根节点
|
|
|
|
|
bst.root = child
|
|
|
|
|
}
|
|
|
|
|
// 子节点数为 2
|
|
|
|
|
} else {
|
|
|
|
|
// 获取中序遍历中待删除节点 cur 的下一个节点
|
|
|
|
|
tmp := cur.Right
|
|
|
|
|
for tmp.Left != nil {
|
|
|
|
|
tmp = tmp.Left
|
|
|
|
|
}
|
|
|
|
|
// 递归删除节点 tmp
|
|
|
|
|
bst.remove(tmp.Val)
|
|
|
|
|
// 用 tmp 覆盖 cur
|
|
|
|
|
cur.Val = tmp.Val
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "JavaScript"
|
|
|
|
|
|
|
|
|
|
```javascript title="binary_search_tree.js"
|
|
|
|
|
/* 删除节点 */
|
|
|
|
|
function remove(num) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (root === null) return;
|
|
|
|
|
let cur = root,
|
|
|
|
|
pre = null;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur !== null) {
|
|
|
|
|
// 找到待删除节点,跳出循环
|
|
|
|
|
if (cur.val === num) break;
|
|
|
|
|
pre = cur;
|
|
|
|
|
// 待删除节点在 cur 的右子树中
|
|
|
|
|
if (cur.val < num) cur = cur.right;
|
|
|
|
|
// 待删除节点在 cur 的左子树中
|
|
|
|
|
else cur = cur.left;
|
|
|
|
|
}
|
|
|
|
|
// 若无待删除节点,则直接返回
|
|
|
|
|
if (cur === null) return;
|
|
|
|
|
// 子节点数量 = 0 or 1
|
|
|
|
|
if (cur.left === null || cur.right === null) {
|
|
|
|
|
// 当子节点数量 = 0 / 1 时, child = null / 该子节点
|
|
|
|
|
let child = cur.left !== null ? cur.left : cur.right;
|
|
|
|
|
// 删除节点 cur
|
|
|
|
|
if (cur != root) {
|
|
|
|
|
if (pre.left === cur) pre.left = child;
|
|
|
|
|
else pre.right = child;
|
|
|
|
|
} else {
|
|
|
|
|
// 若删除节点为根节点,则重新指定根节点
|
|
|
|
|
root = child;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 子节点数量 = 2
|
|
|
|
|
else {
|
|
|
|
|
// 获取中序遍历中 cur 的下一个节点
|
|
|
|
|
let tmp = cur.right;
|
|
|
|
|
while (tmp.left !== null) {
|
|
|
|
|
tmp = tmp.left;
|
|
|
|
|
}
|
|
|
|
|
// 递归删除节点 tmp
|
|
|
|
|
remove(tmp.val);
|
|
|
|
|
// 用 tmp 覆盖 cur
|
|
|
|
|
cur.val = tmp.val;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "TypeScript"
|
|
|
|
|
|
|
|
|
|
```typescript title="binary_search_tree.ts"
|
|
|
|
|
/* 删除节点 */
|
|
|
|
|
function remove(num: number): void {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (root === null) {
|
|
|
|
|
return;
|
|
|
|
|
}
|
|
|
|
|
let cur = root,
|
|
|
|
|
pre: TreeNode | null = null;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur !== null) {
|
|
|
|
|
// 找到待删除节点,跳出循环
|
|
|
|
|
if (cur.val === num) {
|
|
|
|
|
break;
|
|
|
|
|
}
|
|
|
|
|
pre = cur;
|
|
|
|
|
if (cur.val < num) {
|
|
|
|
|
cur = cur.right as TreeNode; // 待删除节点在 cur 的右子树中
|
|
|
|
|
} else {
|
|
|
|
|
cur = cur.left as TreeNode; // 待删除节点在 cur 的左子树中
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 若无待删除节点,则直接返回
|
|
|
|
|
if (cur === null) {
|
|
|
|
|
return;
|
|
|
|
|
}
|
|
|
|
|
// 子节点数量 = 0 or 1
|
|
|
|
|
if (cur.left === null || cur.right === null) {
|
|
|
|
|
// 当子节点数量 = 0 / 1 时, child = null / 该子节点
|
|
|
|
|
let child = cur.left !== null ? cur.left : cur.right;
|
|
|
|
|
// 删除节点 cur
|
|
|
|
|
if (cur != root) {
|
|
|
|
|
if (pre!.left === cur) {
|
|
|
|
|
pre!.left = child;
|
|
|
|
|
} else {
|
|
|
|
|
pre!.right = child;
|
|
|
|
|
}
|
|
|
|
|
} else {
|
|
|
|
|
// 若删除节点为根节点,则重新指定根节点
|
|
|
|
|
root = child;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 子节点数量 = 2
|
|
|
|
|
else {
|
|
|
|
|
// 获取中序遍历中 cur 的下一个节点
|
|
|
|
|
let tmp = cur.right;
|
|
|
|
|
while (tmp.left !== null) {
|
|
|
|
|
tmp = tmp.left;
|
|
|
|
|
}
|
|
|
|
|
// 递归删除节点 tmp
|
|
|
|
|
remove(tmp!.val);
|
|
|
|
|
// 用 tmp 覆盖 cur
|
|
|
|
|
cur.val = tmp.val;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C"
|
|
|
|
|
|
|
|
|
|
```c title="binary_search_tree.c"
|
|
|
|
|
/* 删除节点 */
|
|
|
|
|
// 由于引入了 stdio.h ,此处无法使用 remove 关键词
|
|
|
|
|
void removeNode(binarySearchTree *bst, int num) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (bst->root == NULL)
|
|
|
|
|
return;
|
|
|
|
|
TreeNode *cur = bst->root, *pre = NULL;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur != NULL) {
|
|
|
|
|
// 找到待删除节点,跳出循环
|
|
|
|
|
if (cur->val == num)
|
|
|
|
|
break;
|
|
|
|
|
pre = cur;
|
|
|
|
|
if (cur->val < num) {
|
|
|
|
|
// 待删除节点在 root 的右子树中
|
|
|
|
|
cur = cur->right;
|
|
|
|
|
} else {
|
|
|
|
|
// 待删除节点在 root 的左子树中
|
|
|
|
|
cur = cur->left;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 若无待删除节点,则直接返回
|
|
|
|
|
if (cur == NULL)
|
|
|
|
|
return;
|
|
|
|
|
// 判断待删除节点是否存在子节点
|
|
|
|
|
if (cur->left == NULL || cur->right == NULL) {
|
|
|
|
|
/* 子节点数量 = 0 or 1 */
|
|
|
|
|
// 当子节点数量 = 0 / 1 时, child = nullptr / 该子节点
|
|
|
|
|
TreeNode *child = cur->left != NULL ? cur->left : cur->right;
|
|
|
|
|
// 删除节点 cur
|
|
|
|
|
if (pre->left == cur) {
|
|
|
|
|
pre->left = child;
|
|
|
|
|
} else {
|
|
|
|
|
pre->right = child;
|
|
|
|
|
}
|
|
|
|
|
} else {
|
|
|
|
|
/* 子节点数量 = 2 */
|
|
|
|
|
// 获取中序遍历中 cur 的下一个节点
|
|
|
|
|
TreeNode *tmp = cur->right;
|
|
|
|
|
while (tmp->left != NULL) {
|
|
|
|
|
tmp = tmp->left;
|
|
|
|
|
}
|
|
|
|
|
int tmpVal = tmp->val;
|
|
|
|
|
// 递归删除节点 tmp
|
|
|
|
|
removeNode(bst, tmp->val);
|
|
|
|
|
// 用 tmp 覆盖 cur
|
|
|
|
|
cur->val = tmpVal;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "C#"
|
|
|
|
|
|
|
|
|
|
```csharp title="binary_search_tree.cs"
|
|
|
|
|
/* 删除节点 */
|
|
|
|
|
void remove(int num) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (root == null)
|
|
|
|
|
return;
|
|
|
|
|
TreeNode? cur = root, pre = null;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur != null) {
|
|
|
|
|
// 找到待删除节点,跳出循环
|
|
|
|
|
if (cur.val == num)
|
|
|
|
|
break;
|
|
|
|
|
pre = cur;
|
|
|
|
|
// 待删除节点在 cur 的右子树中
|
|
|
|
|
if (cur.val < num)
|
|
|
|
|
cur = cur.right;
|
|
|
|
|
// 待删除节点在 cur 的左子树中
|
|
|
|
|
else
|
|
|
|
|
cur = cur.left;
|
|
|
|
|
}
|
|
|
|
|
// 若无待删除节点,则直接返回
|
|
|
|
|
if (cur == null || pre == null)
|
|
|
|
|
return;
|
|
|
|
|
// 子节点数量 = 0 or 1
|
|
|
|
|
if (cur.left == null || cur.right == null) {
|
|
|
|
|
// 当子节点数量 = 0 / 1 时, child = null / 该子节点
|
|
|
|
|
TreeNode? child = cur.left != null ? cur.left : cur.right;
|
|
|
|
|
// 删除节点 cur
|
|
|
|
|
if (cur != root) {
|
|
|
|
|
if (pre.left == cur)
|
|
|
|
|
pre.left = child;
|
|
|
|
|
else
|
|
|
|
|
pre.right = child;
|
|
|
|
|
} else {
|
|
|
|
|
// 若删除节点为根节点,则重新指定根节点
|
|
|
|
|
root = child;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 子节点数量 = 2
|
|
|
|
|
else {
|
|
|
|
|
// 获取中序遍历中 cur 的下一个节点
|
|
|
|
|
TreeNode? tmp = cur.right;
|
|
|
|
|
while (tmp.left != null) {
|
|
|
|
|
tmp = tmp.left;
|
|
|
|
|
}
|
|
|
|
|
// 递归删除节点 tmp
|
|
|
|
|
remove(tmp.val);
|
|
|
|
|
// 用 tmp 覆盖 cur
|
|
|
|
|
cur.val = tmp.val;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Swift"
|
|
|
|
|
|
|
|
|
|
```swift title="binary_search_tree.swift"
|
|
|
|
|
/* 删除节点 */
|
|
|
|
|
@discardableResult
|
|
|
|
|
func remove(num: Int) {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if root == nil {
|
|
|
|
|
return
|
|
|
|
|
}
|
|
|
|
|
var cur = root
|
|
|
|
|
var pre: TreeNode?
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while cur != nil {
|
|
|
|
|
// 找到待删除节点,跳出循环
|
|
|
|
|
if cur!.val == num {
|
|
|
|
|
break
|
|
|
|
|
}
|
|
|
|
|
pre = cur
|
|
|
|
|
// 待删除节点在 cur 的右子树中
|
|
|
|
|
if cur!.val < num {
|
|
|
|
|
cur = cur?.right
|
|
|
|
|
}
|
|
|
|
|
// 待删除节点在 cur 的左子树中
|
|
|
|
|
else {
|
|
|
|
|
cur = cur?.left
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 若无待删除节点,则直接返回
|
|
|
|
|
if cur == nil {
|
|
|
|
|
return
|
|
|
|
|
}
|
|
|
|
|
// 子节点数量 = 0 or 1
|
|
|
|
|
if cur?.left == nil || cur?.right == nil {
|
|
|
|
|
// 当子节点数量 = 0 / 1 时, child = null / 该子节点
|
|
|
|
|
let child = cur?.left != nil ? cur?.left : cur?.right
|
|
|
|
|
// 删除节点 cur
|
|
|
|
|
if cur != root {
|
|
|
|
|
if pre?.left === cur {
|
|
|
|
|
pre?.left = child
|
|
|
|
|
} else {
|
|
|
|
|
pre?.right = child
|
|
|
|
|
}
|
|
|
|
|
} else {
|
|
|
|
|
// 若删除节点为根节点,则重新指定根节点
|
|
|
|
|
root = cur;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 子节点数量 = 2
|
|
|
|
|
else {
|
|
|
|
|
// 获取中序遍历中 cur 的下一个节点
|
|
|
|
|
var tmp = cur?.right
|
|
|
|
|
while tmp?.left != nil {
|
|
|
|
|
tmp = tmp?.left
|
|
|
|
|
}
|
|
|
|
|
// 递归删除节点 tmp
|
|
|
|
|
remove(num: tmp!.val)
|
|
|
|
|
// 用 tmp 覆盖 cur
|
|
|
|
|
cur?.val = tmp!.val
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Zig"
|
|
|
|
|
|
|
|
|
|
```zig title="binary_search_tree.zig"
|
|
|
|
|
// 删除节点
|
|
|
|
|
fn remove(self: *Self, num: T) void {
|
|
|
|
|
// 若树为空,直接提前返回
|
|
|
|
|
if (self.root == null) return;
|
|
|
|
|
var cur = self.root;
|
|
|
|
|
var pre: ?*inc.TreeNode(T) = null;
|
|
|
|
|
// 循环查找,越过叶节点后跳出
|
|
|
|
|
while (cur != null) {
|
|
|
|
|
// 找到待删除节点,跳出循环
|
|
|
|
|
if (cur.?.val == num) break;
|
|
|
|
|
pre = cur;
|
|
|
|
|
// 待删除节点在 cur 的右子树中
|
|
|
|
|
if (cur.?.val < num) {
|
|
|
|
|
cur = cur.?.right;
|
|
|
|
|
// 待删除节点在 cur 的左子树中
|
|
|
|
|
} else {
|
|
|
|
|
cur = cur.?.left;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
// 若无待删除节点,则直接返回
|
|
|
|
|
if (cur == null) return;
|
|
|
|
|
// 子节点数量 = 0 or 1
|
|
|
|
|
if (cur.?.left == null or cur.?.right == null) {
|
|
|
|
|
// 当子节点数量 = 0 / 1 时, child = null / 该子节点
|
|
|
|
|
var child = if (cur.?.left != null) cur.?.left else cur.?.right;
|
|
|
|
|
// 删除节点 cur
|
|
|
|
|
if (pre.?.left == cur) {
|
|
|
|
|
pre.?.left = child;
|
|
|
|
|
} else {
|
|
|
|
|
pre.?.right = child;
|
|
|
|
|
}
|
|
|
|
|
// 子节点数量 = 2
|
|
|
|
|
} else {
|
|
|
|
|
// 获取中序遍历中 cur 的下一个节点
|
|
|
|
|
var tmp = cur.?.right;
|
|
|
|
|
while (tmp.?.left != null) {
|
|
|
|
|
tmp = tmp.?.left;
|
|
|
|
|
}
|
|
|
|
|
var tmp_val = tmp.?.val;
|
|
|
|
|
// 递归删除节点 tmp
|
|
|
|
|
self.remove(tmp.?.val);
|
|
|
|
|
// 用 tmp 覆盖 cur
|
|
|
|
|
cur.?.val = tmp_val;
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
=== "Dart"
|
|
|
|
|
|
|
|
|
|
```dart title="binary_search_tree.dart"
|
|
|
|
|
[class]{BinarySearchTree}-[func]{remove}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
### 排序
|
|
|
|
|
|
|
|
|
|
我们知道,二叉树的中序遍历遵循“左 $\rightarrow$ 根 $\rightarrow$ 右”的遍历顺序,而二叉搜索树满足“左子节点 $<$ 根节点 $<$ 右子节点”的大小关系。因此,在二叉搜索树中进行中序遍历时,总是会优先遍历下一个最小节点,从而得出一个重要性质:**二叉搜索树的中序遍历序列是升序的**。
|
|
|
|
|
|
|
|
|
|
利用中序遍历升序的性质,我们在二叉搜索树中获取有序数据仅需 $O(n)$ 时间,无需额外排序,非常高效。
|
|
|
|
|
|
|
|
|
|
![二叉搜索树的中序遍历序列](binary_search_tree.assets/bst_inorder_traversal.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 二叉搜索树的中序遍历序列 </p>
|
|
|
|
|
|
|
|
|
|
## 7.4.2. 二叉搜索树的效率
|
|
|
|
|
|
|
|
|
|
给定一组数据,我们考虑使用数组或二叉搜索树存储。
|
|
|
|
|
|
|
|
|
|
观察可知,二叉搜索树的各项操作的时间复杂度都是对数阶,具有稳定且高效的性能表现。只有在高频添加、低频查找删除的数据适用场景下,数组比二叉搜索树的效率更高。
|
|
|
|
|
|
|
|
|
|
<div class="center-table" markdown>
|
|
|
|
|
|
|
|
|
|
| | 无序数组 | 二叉搜索树 |
|
|
|
|
|
| -------- | -------- | ----------- |
|
|
|
|
|
| 查找元素 | $O(n)$ | $O(\log n)$ |
|
|
|
|
|
| 插入元素 | $O(1)$ | $O(\log n)$ |
|
|
|
|
|
| 删除元素 | $O(n)$ | $O(\log n)$ |
|
|
|
|
|
|
|
|
|
|
</div>
|
|
|
|
|
|
|
|
|
|
在理想情况下,二叉搜索树是“平衡”的,这样就可以在 $\log n$ 轮循环内查找任意节点。
|
|
|
|
|
|
|
|
|
|
然而,如果我们在二叉搜索树中不断地插入和删除节点,可能导致二叉树退化为链表,这时各种操作的时间复杂度也会退化为 $O(n)$ 。
|
|
|
|
|
|
|
|
|
|
![二叉搜索树的平衡与退化](binary_search_tree.assets/bst_degradation.png)
|
|
|
|
|
|
|
|
|
|
<p align="center"> Fig. 二叉搜索树的平衡与退化 </p>
|
|
|
|
|
|
|
|
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## 7.4.3. 二叉搜索树常见应用
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- 用作系统中的多级索引,实现高效的查找、插入、删除操作。
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- 作为某些搜索算法的底层数据结构。
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- 用于存储数据流,以保持其有序状态。
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