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@ -445,19 +445,15 @@ comments: true
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3. 使用 `nex` 替换待删除结点;
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=== "Step 1"
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![bst_remove_case3_1](binary_search_tree.assets/bst_remove_case3_1.png)
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=== "Step 2"
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![bst_remove_case3_2](binary_search_tree.assets/bst_remove_case3_2.png)
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=== "Step 3"
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![bst_remove_case3_3](binary_search_tree.assets/bst_remove_case3_3.png)
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=== "Step 4"
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![bst_remove_case3_4](binary_search_tree.assets/bst_remove_case3_4.png)
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删除结点操作也使用 $O(\log n)$ 时间,其中查找待删除结点 $O(\log n)$ ,获取中序遍历后继结点 $O(\log n)$ 。
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@ -489,11 +485,13 @@ comments: true
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// 删除结点 cur
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if (pre.left == cur) pre.left = child;
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else pre.right = child;
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// 释放内存
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delete cur;
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}
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// 子结点数量 = 2
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else {
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// 获取中序遍历中 cur 的下一个结点
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TreeNode nex = min(cur.right);
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TreeNode nex = getInOrderNext(cur.right);
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int tmp = nex.val;
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// 递归删除结点 nex
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remove(nex.val);
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@ -502,15 +500,6 @@ comments: true
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}
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return cur;
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}
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/* 获取最小结点 */
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TreeNode min(TreeNode root) {
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if (root == null) return root;
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// 循环访问左子结点,直到叶结点时为最小结点,跳出
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while (root.left != null) {
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root = root.left;
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}
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return root;
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}
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```
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=== "C++"
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@ -544,7 +533,7 @@ comments: true
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// 子结点数量 = 2
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else {
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// 获取中序遍历中 cur 的下一个结点
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TreeNode* nex = min(cur->right);
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TreeNode* nex = getInOrderNext(cur->right);
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int tmp = nex->val;
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// 递归删除结点 nex
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remove(nex->val);
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@ -553,15 +542,6 @@ comments: true
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}
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return cur;
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}
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/* 获取最小结点 */
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TreeNode* min(TreeNode* root) {
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if (root == nullptr) return root;
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// 循环访问左子结点,直到叶结点时为最小结点,跳出
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while (root->left != nullptr) {
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root = root->left;
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}
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return root;
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}
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```
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=== "Python"
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@ -604,23 +584,13 @@ comments: true
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# 子结点数量 = 2
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else:
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# 获取中序遍历中 cur 的下一个结点
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nex = self.min(cur.right)
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nex = self.get_inorder_next(cur.right)
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tmp = nex.val
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# 递归删除结点 nex
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self.remove(nex.val)
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# 将 nex 的值复制给 cur
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cur.val = tmp
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return cur
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""" 获取最小结点 """
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def min(self, root: typing.Optional[TreeNode]) -> typing.Optional[TreeNode]:
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if root is None:
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return root
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# 循环访问左子结点,直到叶结点时为最小结点,跳出
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while root.left is not None:
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root = root.left
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return root
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```
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=== "Go"
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@ -671,7 +641,7 @@ comments: true
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// 子结点数为 2
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} else {
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// 获取中序遍历中待删除结点 cur 的下一个结点
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next := bst.GetInorderNext(cur)
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next := bst.GetInOrderNext(cur)
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temp := next.Val
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// 递归删除结点 next
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bst.Remove(next.Val)
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@ -713,7 +683,7 @@ comments: true
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// 子结点数量 = 2
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else {
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// 获取中序遍历中 cur 的下一个结点
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let nex = min(cur.right);
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let nex = getInOrderNext(cur.right);
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let tmp = nex.val;
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// 递归删除结点 nex
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remove(nex.val);
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@ -766,7 +736,7 @@ comments: true
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// 子结点数量 = 2
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else {
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// 获取中序遍历中 cur 的下一个结点
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let next = min(cur.right);
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let next = getInOrderNext(cur.right);
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let tmp = next!.val;
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// 递归删除结点 nex
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remove(next!.val);
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@ -824,7 +794,7 @@ comments: true
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else
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{
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// 获取中序遍历中 cur 的下一个结点
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TreeNode? nex = min(cur.right);
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TreeNode? nex = getInOrderNext(cur.right);
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if (nex != null)
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{
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int tmp = nex.val;
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@ -836,18 +806,6 @@ comments: true
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}
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return cur;
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}
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/* 获取最小结点 */
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TreeNode? min(TreeNode? root)
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{
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if (root == null) return root;
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// 循环访问左子结点,直到叶结点时为最小结点,跳出
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while (root.left != null)
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{
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root = root.left;
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}
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return root;
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}
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```
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## 二叉搜索树的优势
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