Binary Tree Postorder Traversal Given a binary tree, return the postorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? 解法一:递归法 /**…
Problem Link: http://oj.leetcode.com/problems/binary-tree-postorder-traversal/ The post-order-traversal of a binary tree is a classic problem, the recursive way to solve it is really straightforward, the pseudo-code is as follows. RECURSIVE-POST-ORDE…
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 迭代 日期 题目地址:https://leetcode.com/problems/binary-tree-postorder-traversal/ 题目描述 Given a binary tree, return the postorder traversal of its nodes' values. Example: Input: [1,nu…
Given a binary tree, return the postorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? 1.递归算法的递归定义是: 若二叉树为空,则遍历结束:否则⑴ 后序遍历左子树(递归调用…
Given a binary tree, return the postorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? Subscribe to see which companies asked this…
Given a binary tree, return the postorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? Subscribe to see which companies asked this…
这是悦乐书的第269次更新,第283篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第136题(顺位题号是590).给定一个n-ary树,返回其节点值的后序遍历.例如,给定一个3-ary树: 1 / | \ 3 2 4 / \ 5 6 其后序遍历结果为:[5,6,3,2,4,1]. 注意:递归解决方案是微不足道的,你可以用迭代的方式做吗? 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试. 02…
Difficulty: Hard  More:[目录]LeetCode Java实现 Description https://leetcode.com/problems/binary-tree-postorder-traversal/ Given a binary tree, return the postordertraversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [3,2,1] Foll…
1.题目描述 2.问题分析 递归. 3.代码 vector<int> postorder(Node* root) { vector<int> v; postNorder(root, v); return v; } void postNorder(Node *root, vector<int> &v) { if (root == NULL) return; for (auto it = root->children.begin(); it != root-&…
1.题目描述 2.问题分析 递归 3.代码 vector<int> postorderTraversal(TreeNode* root) { vector<int> v; postBorder(root,v); return v; } void postBorder(TreeNode *root, vector<int> &v) { if (root == NULL) return ; postBorder(root->left, v); postBord…