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. 题解: 采用递归的方法,要记录depth用来比较. 代码如下:…

Question 110. Balanced Binary Tree Solution 题目大意:判断一个二叉树是不是平衡二叉树 思路:定义个boolean来记录每个子节点是否平衡 Java实现: public boolean isBalanced(TreeNode root) { boolean[] balanced = {true}; height(root, balanced); return balanced[0]; } private int height(TreeNode node,…

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. 很早以前做的了  准…

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. /** * Definition for 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. Solution:  bool helperBalanced(int…

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:给定一个二叉树,判断是否是高度平衡的,这里的高度平衡定义是每个节…

题目: Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 题解: 递归解法急速判断左右两边子树哪个depth最小,要注意如果有个节点只有一边孩子时,不能返回0,要返回另外一半边的depth. 递归解法:              …

题目: Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 题解: 递归解法:                                                            } 非递归解法,参考codegan…

这是悦乐书的第167次更新,第169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26题(顺位题号是110).给定二叉树,判断它是否是高度平衡的.对于此问题,高度平衡二叉树定义为:一个二叉树,其中每个节点的两个子树的深度从不相差超过1.例如: 给定以下树[3,9,20,null,null,15,7]: 3 / \ 9 20 / \ 15 7 返回true. 给定以下树[1,2,2,3,3,null,null,4,4]: 1 / \ 2 2 / \ 3 3 / \…

递归 左子树是否为平衡二叉树 右子树是否为平衡二叉树 左右子树高度差是否大于1,大于1,返回false 否则判断左右子树 最简单的理解方法就是如下的思路: class Solution { public: bool isBalanced(TreeNode* root) { if(root==NULL){ return true; } int ldepth=depth(root->left); int rdepth=depth(root->right); )return false; else{…