LeetCode (65):Same tree】的更多相关文章

Total Accepted: 83663 Total Submissions: 200541 Difficulty: Easy Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 首…
// 我的代码 package Leetcode; /** * 199. Binary Tree Right Side View * address: https://leetcode.com/problems/binary-tree-right-side-view/ * Given a binary tree, imagine yourself standing on the right side of it, * return the values of the nodes you can…
判断一棵树是否自对称 可以回忆我们做过的Leetcode 100 Same Tree 二叉树和Leetcode 226 Invert Binary Tree 二叉树 先可以将左子树进行Invert Binary Tree,然后用Same Tree比较左右子树 而我的做法是改下Same Tree的函数,改动的是第27行 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left;…
LeetCode:Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree.                                            …
Given a binary tree, return the postorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [3,2,1] Follow up: Recursive solution is trivial, could you do it iteratively? --------------------------------------------------…
版权声明: 本文为博主Bravo Yeung(知乎UserName同名)的原创文章,欲转载请先私信获博主允许,转载时请附上网址 http://blog.csdn.net/lzuacm. Leetcode 65. 有效数字 Leetcode 65. Valid Number 在线提交: Leetcode https://leetcode.com/problems/valid-number/ 类似问题 - PAT 1014_牛客网 https://www.nowcoder.com/pat/6/pro…
leetcode 199. Binary Tree Right Side View 这个题实际上就是把每一行最右侧的树打印出来,所以实际上还是一个层次遍历. 依旧利用之前层次遍历的代码,每次大的循环存储的是一行的节点,最后一个节点就是想要的那个节点 class Solution { public: vector<int> rightSideView(TreeNode* root) { vector<int> result; if(root == NULL) return resul…
Given a binary tree, you need to find the length of Longest Consecutive Path in Binary Tree. Especially, this path can be either increasing or decreasing. For example, [1,2,3,4] and [4,3,2,1] are both considered valid, but the path [1,2,4,3] is not v…
Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example, given a 3-ary tree: We should return its level order traversal: [ [1], [3,2,4], [5,6] ] Note: The depth of the tree is…
翻译 给定一个二叉树.返回其兴许遍历的节点的值. 比如: 给定二叉树为 {1. #, 2, 3} 1 \ 2 / 3 返回 [3, 2, 1] 备注:用递归是微不足道的,你能够用迭代来完毕它吗? 原文 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: R…