【LeetCode】#145二叉树的后序遍历(Binary Tree Postorder Traversal) 题目描述 给定一个二叉树,返回它的 后序 遍历。 示例 输入: [1,null,2,3] 输出: [3,2,1] Description Given a binary tree, return the postorder traversal of its nodes’ val...leet...
代码如下: 1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12vector<int> postorderTraversal(TreeNode *root) {13//IMPORTANT: Please...
* int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> postorderTraversal(TreeNode *root) { vector<int>perm;if(!root)returnperm; stack<TreeNode *>sta; TreeNode*p = root, *pre =NULL...
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? 本题难度Hard。有3种算法分别是: 递归...
Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 代码语言:javascript 代码运行次数:0 1\2/3 return[3,2,1]. Note: Recursive solution is trivial, could you do it iteratively?
private void pushAllTheLeft(Stack<TreeNode> s, TreeNode root){ s.push(root); while(root.left!=null){ root = root.left; s.push(root); } } } Binary Tree Postorder Traversal Given a binary tree, return the postorder traversal of its nodes' values. ...
Leetcode 之Binary Tree Postorder Traversal(44) 后序遍历,比先序和中序都要复杂。访问一个结点前,需要先判断其右孩子是否被访问过。如果是,则可以访问该结点;否则,需要先处理右子树。 vector<int> postorderTraversal(TreeNode *root) { vector<int>result;...
1. Problem Descriptions:Given two integer arrays inorderandpostorderwhereinorderis the inorder traversal of a binary tree andpostorderis the postorder traversal of the same tree, construct and retu…
/** * Definition for a binary tree node. * function TreeNode(val) { * this.val = val; * this.left = this.right = null; * } *//** * @param {TreeNode} root * @return {number[]} */var postorderTraversal = function (root) { // 1. Recursive solution // if (!roo...
left); preorderTraversal(root.right); } return res; } } 迭代法: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List<Integer> preorder...