专业开发网站设计,外贸开发网站公司,学做缝纫的网站,门户网站建设验收报告题目链接#xff1a;110. 平衡二叉树#xff08;简单#xff09; 算法原理#xff1a; 解法#xff1a;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#x1f447;目录位置#xff1a;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…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(rootnull) return true; int lefthightgethight(root.left); int righthightgethight(root.right); return Math.abs(lefthight-righthight)2isBalanced(root.left)isBalanced(root.right); } private int gethight(TreeNode root){ if(rootnull) return 0; int lefthightgethight(root.left); int righthightgethight(root.right); return lefthightrighthight?lefthight1:righthight1; } }