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题目：输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。

/** * 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 {
       // 封装一个返回类    public static class ReturnData{
           private boolean isB ;        private int h;        public ReturnData(boolean isB,int h){
               this.isB = isB;            this.h = h;        }    }        // 迭代判断    public static ReturnData process(TreeNode root){
           if(root==null){
               return new ReturnData(true,0);        }        ReturnData leftData = process(root.left);        if(!leftData.isB){
               return new ReturnData(false,0);        }        ReturnData rightData = process(root.right);        if(!rightData.isB){
               return new ReturnData(false,0);        }        if(Math.abs(rightData.h-leftData.h)>1){
               return new ReturnData(false,0);        }        return new ReturnData(true,Math.max(rightData.h,leftData.h)+1);    }        public boolean isBalanced(TreeNode root) {
           return process(root).isB;    }}

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