要求:判断一棵树是否是平衡二叉树 Given a binary tree, determine if it is height-balanced. For . 代码如下: struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x): val(x),left(NULL), right(NULL) {} }; int maxTreeDepth(TreeNode *root) //先求树的深度 { if (NU…

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 求二叉树是否平衡,根据题目中的定义,高度平衡二叉树是每一个节点的两个字…

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 题意:二叉树中以任意节点为父结点的两棵子树的深度差不超过1,为平衡二叉…

Given a binary tree, return all root-to-leaf paths. Note: A leaf is a node with no children. Example: Input: 1 / \ 2 3 \ 5 Output: ["1->2->5", "1->3"] Explanation: All root-to-leaf paths are: 1->2->5, 1->3 思路为DFS, 只是a…

4.1 Implement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one. LeetCode上的原题,请参见我之前的博客Bala…

// 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…

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? --------------------------------------------------…

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…