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.
题意:
给定一棵二叉树,判定它是否为平衡二叉树。
算法分析:
* 平衡二叉树(Balanced Binary Tree)又被称为AVL树(有别于AVL算法),
* 且具有以下性质:它是一棵空树或它的左右两个子树的高度差的绝对值不超过1,
* 并且左右两个子树都是一棵平衡二叉树。
* 下面的代码就完全按照定义,首先得到节点左右子树的高度(递归),然后判断左右子树是否为平衡二叉树,
* 利用递归完成整棵树的判断。完全满足条件,就返回true.
AC代码:
public class Solution { //递归 public boolean isBalanced(TreeNode root) { if(root==null) return true; int leftdepth; int rightdepth; leftdepth=maxDepth(root.left); rightdepth=maxDepth(root.right); if(Math.abs(leftdepth-rightdepth)<=1&&isBalanced(root.left)&&isBalanced(root.right)) return true; else return false; } //又是递归 采用代码《Maximum Depth of Binary Tree 》--https://leetcode.com/problems/maximum-depth-of-binary-tree/ public int maxDepth(TreeNode root) { if(root == null) { return 0; } int l = maxDepth(root.left); int r = maxDepth(root.right); return l > r? l + 1:r+ 1; } }
时间: 2024-11-10 19:16:23