剑指offer 面试题39:判断平衡二叉树 提交网址:  http://www.nowcoder.com/practice/8b3b95850edb4115918ecebdf1b4d222?tpId=13&tqId=11192 时间限制:1秒       空间限制:32768K      参与人数:2481 题目描述 输入一棵二叉树,判断该二叉树是否是平衡二叉树. 分析: 平衡二叉树定义 递归解法 AC代码: #include<iostream> #include<vector&…

110. 平衡二叉树 110. Balanced Binary Tree 题目描述 给定一个二叉树,判断它是否是高度平衡的二叉树. 本题中,一棵高度平衡二叉树定义为: 一个二叉树每个节点的左右两个子树的高度差的绝对值不超过 1. 每日一算法2019/5/18Day 15LeetCode110. Balanced Binary Tree 示例 1: 给定二叉树 [3,9,20,null,null,15,7] 3 / \ 9 20 / \ 15 7 返回 true. 示例 2: 给定二叉树 [1,2…

判断一棵树是否是平衡树,即左右子树的深度相差不超过1. 我们可以回顾下depth函数其实是Leetcode 104 Maximum Depth of Binary Tree 二叉树 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL…

110. Balanced Binary Tree 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. 判断二叉树是否为平…

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. 题目标签:Tree 这道题目给了我们一个 二叉树, 让我们来判断一下它…

110. Balanced Binary Tree 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. Example…

这篇文章介绍Leetcode1到10题的解决思路和相关代码. 1. Two sum 问题描述:给定一个整数数组,返回两个数字的索引,使它们加起来等于一个特定的目标. 例子: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. 常规方法:使用双重循环,第一重从左往右固定索引,计算需要查找的结果,第二层循环从固定索引出发依次向右查找第一层计算的结果.时间复杂度\(…

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. Example 1: Given the following tre…

110. 平衡二叉树 给定一个二叉树,判断它是否是高度平衡的二叉树. 本题中,一棵高度平衡二叉树定义为: 一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过1. 示例 1: 给定二叉树 [3,9,20,null,null,15,7] 3 / \ 9 20 / \ 15 7 返回 true . 示例 2: 给定二叉树 [1,2,2,3,3,null,null,4,4] 1 / \ 2 2 / \ 3 3 / \ 4 4 返回 false . PS: 模版一共三步,就是递归的三部曲: 找终止条…

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. def isBalanced(self, root): ""…

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. 判断一棵树是否是平衡二叉树 利用高度的答案,递归求解. /** * D…

题目描述: 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. 解题思路: 递归法解题. 代码如下: /** * Defi…

Problem 1 [Balanced Binary Tree] 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. Pr…

问题 给出一棵二叉树,判断它是否在高度上是平衡的. 对于本问题,高度上平衡的二叉树定义为:每个节点的两棵子树的深度差永远不大于1的一棵二叉树. 初始思路 根据定义,思路应该比较直接:递归计算每个节点左右子树的深度,只要发现一次深度差大于1的情况,即可终止递归返回不平衡的结果.最终代码如下: class Solution { public: bool isBalanced(TreeNode *root) { if(!root) { return true; } isBalanced_ = true…

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. 问题:给定一个二叉树,判断其是否平衡. 当二叉树中所有节点的左右子树高…

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. Example 1: Given the following tre…

Balanced Binary Tree [数据结构和算法]全面剖析树的各类遍历方法 描述 解析 递归分别判断每个节点的左右子树 该题是Easy的原因是该题可以很容易的想到时间复杂度为O(n^2)的方法.即按照定义,判断根节点左右子树的高度是不是相差1,递归判断左右子树是不是平衡的. 根据深度判断左右子树是否平衡 在计算树的高度的同时判断该树是不是平衡的. 即,先判断子树是不是平衡的,若是,则返回子树的高度:若不是,则返回一个非法的数字,如负数. 当一个节点是左右子树有一个不是平衡二叉树则不必继…

翻译 给定一个二叉树,决定它是否是高度平衡的. (高度是名词不是形容词-- 对于这个问题.一个高度平衡二叉树被定义为: 这棵树的每一个节点的两个子树的深度差不能超过1. 原文 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…

给定一个二叉树,判断它是否是高度平衡的二叉树. 本题中,一棵高度平衡二叉树定义为: 一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过1. 示例 1: 给定二叉树 [3,9,20,null,null,15,7] 3 / \ 9 20 / \ 15 7 返回 true . 示例 2: 给定二叉树 [1,2,2,3,3,null,null,4,4] 1 / \ 2 2 / \ 3 3 / \ 4 4 返回 false . #include <iostream> #include <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. 解题思路: 递归即可,JAVA实现如下: public boolean…

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:给一棵二叉树,判断它是否是平衡二叉树(即每个节点的左右子树深度之…

原题链接 水题 深度搜索每一节点的左右深度,左右深度差大于1就返回false. class Solution { public: bool isBalanced(TreeNode *root) { bool flag = true; if (!root) return true; dfs(root, flag); return flag; } private: int dfs(TreeNode *root, bool &flag) { if (!root) return 0; int leftH…

题目: 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. Example 1: Given the following…

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. Example 1: Given the following tre…

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int height(TreeNode root){ if(root == null) return 0; int left_height = hei…

参考[LeetCode] questions conlusion_InOrder, PreOrder, PostOrder traversal 可以对binary tree进行遍历. 此处说明Divide and Conquer 的做法,其实跟recursive的做法很像,但是将结果存进array并且输出,最后conquer (这一步worst T:O(n)) 起来,所以时间复杂度可以从遍历O(n) -> O(n^2). 实际上代码是一样, 就是把[root.val] 放在先, 中, 后就是pr…

老年退役选手的 LeetCode 休闲之旅 前言 不知不觉两年多的大学时光悄然流逝,浑浑噩噩的状态似乎从来没有离开过自己. 这两年刷题似乎一直是常态.在退役之后的现在,深感有些东西一段时间没有接触,很容易就变得陌生,遂萌生了刷 LeetCode 的想法,不知这一次能维持多久,谨以此记录来不时地警醒自己. -- 记于2018年寒假 LeetCode之旅 一开始的计划是先不考虑tag,把题目限制在 1~200 题,刷题的顺序是按照难度(Easy-Medium-Hard),题号(1~200). 结果一…

Note: 后面数字n表明刷的第n + 1遍, 如果题目有**, 表明有待总结 Conclusion questions: [LeetCode] questions conclustion_BFS, DFS LeetCode questions conclustion_Path in Tree [LeetCode] questions conlusion_InOrder, PreOrder, PostOrder traversal [LeetCode] questions for Dynamic…

BFS, DFS 的题目总结. Directed graph: Directed Graph Loop detection and if not have, path to print all path. BFS/DFS: (可以用BFS或者DFS的,主要还是遍历) [LeetCode] 733. Flood Fill_Easy tag: BFS     1 [LeetCode] 690. Employee Importance_Easy tag: BFS    1 [LeetCode] 529…

算法思想 二分查找 贪心思想 双指针 排序 快速选择 堆排序 桶排序 搜索 BFS DFS Backtracking 分治 动态规划 分割整数 矩阵路径 斐波那契数列 最长递增子序列 最长公共子系列 0-1 背包 数组区间 字符串编辑 其它问题 数学 素数 最大公约数 进制转换 阶乘 字符串加法减法 相遇问题 多数投票问题 其它 数据结构相关 栈和队列 哈希表 字符串 数组与矩阵 1-n 分布 有序矩阵 链表 树 递归 层次遍历 前中后序遍历 BST Trie 图 位运算 参考资料 算法思想 二…