Rewrite the tree serialization and deserialization methods. Add applications of array and linked list.pull/601/head^2
krahets
1 year ago
parent
c68f18e480
commit
4e13755023
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/**
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* File: array_binary_tree.cpp
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* Created Time: 2023-07-19
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* Author: Krahets (krahets@163.com)
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*/
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#include "../utils/common.hpp"
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/* 数组表示下的二叉树类 */
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class ArrayBinaryTree {
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public:
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/* 构造方法 */
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ArrayBinaryTree(vector<int> arr) {
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tree = arr;
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}
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/* 节点数量 */
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int size() {
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return tree.size();
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}
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/* 获取索引为 i 节点的值 */
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int val(int i) {
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// 若索引越界，则返回 INT_MAX ，代表空位
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if (i < 0 || i >= size())
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return INT_MAX;
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return tree[i];
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}
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/* 获取索引为 i 节点的左子节点的索引 */
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int left(int i) {
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return 2 * i + 1;
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}
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/* 获取索引为 i 节点的右子节点的索引 */
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int right(int i) {
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return 2 * i + 2;
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}
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/* 获取索引为 i 节点的父节点的索引 */
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int parent(int i) {
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return (i - 1) / 2;
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}
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/* 层序遍历 */
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vector<int> levelOrder() {
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vector<int> res;
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// 直接遍历数组
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for (int i = 0; i < size(); i++) {
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if (val(i) != INT_MAX)
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res.push_back(val(i));
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}
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return res;
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}
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/* 前序遍历 */
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vector<int> preOrder() {
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vector<int> res;
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dfs(0, "pre", res);
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return res;
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}
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/* 中序遍历 */
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vector<int> inOrder() {
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vector<int> res;
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dfs(0, "in", res);
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return res;
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}
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/* 后序遍历 */
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vector<int> postOrder() {
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vector<int> res;
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dfs(0, "post", res);
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return res;
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}
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private:
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vector<int> tree;
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/* 深度优先遍历 */
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void dfs(int i, string order, vector<int> &res) {
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// 若为空位，则返回
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if (val(i) == INT_MAX)
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return;
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// 前序遍历
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if (order == "pre")
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res.push_back(val(i));
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dfs(left(i), order, res);
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// 中序遍历
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if (order == "in")
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res.push_back(val(i));
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dfs(right(i), order, res);
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// 后序遍历
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if (order == "post")
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res.push_back(val(i));
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}
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};
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/* Driver Code */
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int main() {
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// 初始化二叉树
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// 使用 INT_MAX 代表空位 nullptr
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vector<int> arr = {1, 2, 3, 4, INT_MAX, 6, 7, 8, 9, INT_MAX, INT_MAX, 12, INT_MAX, INT_MAX, 15};
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TreeNode *root = vectorToTree(arr);
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cout << "\n初始化二叉树\n";
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cout << "二叉树的数组表示：\n";
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printVector(arr);
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cout << "二叉树的链表表示：\n";
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printTree(root);
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// 数组表示下的二叉树类
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ArrayBinaryTree abt(arr);
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// 访问节点
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int i = 1;
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int l = abt.left(i), r = abt.right(i), p = abt.parent(i);
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cout << "\n当前节点的索引为 " << i << "，值为 " << abt.val(i) << "\n";
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cout << "其左子节点的索引为 " << l << "，值为 " << (l != INT_MAX ? to_string(abt.val(l)) : "None") << "\n";
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cout << "其右子节点的索引为 " << r << "，值为 " << (r != INT_MAX ? to_string(abt.val(r)) : "None") << "\n";
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cout << "其父节点的索引为 " << p << "，值为 " << (p != INT_MAX ? to_string(abt.val(p)) : "None") << "\n";
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// 遍历树
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vector<int> res = abt.levelOrder();
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cout << "\n层序遍历为： ";
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printVector(res);
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res = abt.preOrder();
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cout << "前序遍历为： ";
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printVector(res);
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res = abt.inOrder();
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cout << "中序遍历为： ";
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printVector(res);
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res = abt.postOrder();
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cout << "后序遍历为： ";
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printVector(res);
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return 0;
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}
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@ -0,0 +1,118 @@
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"""
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File: array_binary_tree.py
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Created Time: 2023-07-19
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Author: Krahets (krahets@163.com)
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"""
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import sys, os.path as osp
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sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__))))
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from modules import *
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class ArrayBinaryTree:
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"""数组表示下的二叉树类"""
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def __init__(self, arr: list[int | None]):
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"""构造方法"""
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self.__tree = list(arr)
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def size(self):
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"""节点数量"""
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return len(self.__tree)
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def val(self, i: int) -> int:
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"""获取索引为 i 节点的值"""
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# 若索引越界，则返回 None ，代表空位
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if i < 0 or i >= self.size():
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return None
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return self.__tree[i]
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def left(self, i: int) -> int | None:
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"""获取索引为 i 节点的左子节点的索引"""
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return 2 * i + 1
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def right(self, i: int) -> int | None:
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"""获取索引为 i 节点的右子节点的索引"""
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return 2 * i + 2
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def parent(self, i: int) -> int | None:
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"""获取索引为 i 节点的父节点的索引"""
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return (i - 1) // 2
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def level_order(self) -> list[int]:
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"""层序遍历"""
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self.res = []
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# 直接遍历数组
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for i in range(self.size()):
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if self.val(i) is not None:
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self.res.append(self.val(i))
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return self.res
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def __dfs(self, i: int, order: str):
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"""深度优先遍历"""
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if self.val(i) is None:
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return
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# 前序遍历
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if order == "pre":
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self.res.append(self.val(i))
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self.__dfs(self.left(i), order)
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# 中序遍历
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if order == "in":
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self.res.append(self.val(i))
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self.__dfs(self.right(i), order)
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# 后序遍历
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if order == "post":
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self.res.append(self.val(i))
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def pre_order(self) -> list[int]:
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"""前序遍历"""
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self.res = []
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self.__dfs(0, order="pre")
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return self.res
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def in_order(self) -> list[int]:
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"""中序遍历"""
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self.res = []
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self.__dfs(0, order="in")
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return self.res
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def post_order(self) -> list[int]:
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"""后序遍历"""
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self.res = []
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self.__dfs(0, order="post")
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return self.res
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"""Driver Code"""
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if __name__ == "__main__":
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# 初始化二叉树
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# 这里借助了一个从数组直接生成二叉树的函数
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arr = [1, 2, 3, 4, None, 6, 7, 8, 9, None, None, 12, None, None, 15]
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root = list_to_tree(arr)
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print("\n初始化二叉树\n")
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print(f"二叉树的数组表示：")
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print(arr)
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print(f"二叉树的链表表示：")
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print_tree(root)
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# 数组表示下的二叉树类
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abt = ArrayBinaryTree(arr)
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# 访问节点
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i = 1
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l, r, p = abt.left(i), abt.right(i), abt.parent(i)
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print(f"\n当前节点的索引为 {i} ，值为 {abt.val(i)}")
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print(f"其左子节点的索引为 {l} ，值为 {abt.val(l)}")
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print(f"其右子节点的索引为 {r} ，值为 {abt.val(r)}")
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print(f"其父节点的索引为 {p} ，值为 {abt.val(p)}")
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# 遍历树
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res = abt.level_order()
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print("\n层序遍历为：", res)
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res = abt.pre_order()
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print("前序遍历为：", res)
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res = abt.in_order()
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print("中序遍历为：", res)
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res = abt.post_order()
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print("后序遍历为：", res)
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@ -1,72 +0,0 @@
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"""
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File: binary_tree.py
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Created Time: 2021-12-11
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Author: Krahets (krahets@163.com)
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"""
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from collections import deque
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class TreeNode:
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"""Definition for a binary tree node"""
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def __init__(self, val: int = 0):
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self.val: int = val # 节点值
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self.height: int = 0 # 节点高度
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self.left: TreeNode | None = None # 左子节点引用
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self.right: TreeNode | None = None # 右子节点引用
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def list_to_tree(arr: list[int]) -> TreeNode | None:
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"""Generate a binary tree with a list"""
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if not arr:
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return None
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i = 0
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root = TreeNode(arr[0])
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queue: deque[TreeNode] = deque([root])
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while queue:
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node: TreeNode = queue.popleft()
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i += 1
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if i >= len(arr):
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break
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if arr[i] != None:
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node.left = TreeNode(arr[i])
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queue.append(node.left)
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i += 1
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if i >= len(arr):
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break
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if arr[i] != None:
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node.right = TreeNode(arr[i])
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queue.append(node.right)
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return root
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def tree_to_list(root: TreeNode | None) -> list[int]:
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"""Serialize a tree into an array"""
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if not root:
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return []
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queue: deque[TreeNode] = deque()
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queue.append(root)
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res: list[int] = []
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while queue:
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node: TreeNode | None = queue.popleft()
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if node:
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res.append(node.val)
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queue.append(node.left)
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queue.append(node.right)
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else:
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res.append(None)
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return res
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def get_tree_node(root: TreeNode | None, val: int) -> TreeNode | None:
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"""Get a tree node with specific value in a binary tree"""
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if not root:
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return
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if root.val == val:
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return root
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left: TreeNode | None = get_tree_node(root.left, val)
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right: TreeNode | None = get_tree_node(root.right, val)
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return left if left else right
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@ -1,5 +1,5 @@
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"""
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File: linked_list.py
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File: list_node.py
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Created Time: 2021-12-11
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Author: Krahets (krahets@163.com)
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"""
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