A.每日一题——110. 平衡二叉树

题目链接：110. 平衡二叉树（简单）

算法原理：

解法：递归

1ms击败71.53%

时间复杂度O(N)

下面这篇博客有详细解析👇目录位置：OJ面试题→判断平衡二叉树

Java数据结构——7.二叉树《干货笔记》

Java代码：

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public boolean isBalanced(TreeNode root) { //空树也是平衡二叉树 if(root==null) return true; int lefthight=gethight(root.left); int righthight=gethight(root.right); return Math.abs(lefthight-righthight)<2&&isBalanced(root.left)&&isBalanced(root.right); } private int gethight(TreeNode root){ if(root==null) return 0; int lefthight=gethight(root.left); int righthight=gethight(root.right); return lefthight>righthight?lefthight+1:righthight+1; } }