剑指Offer打卡29 —— AcWing 72. 平衡二叉树

【题目描述 】

AcWing 72. 平衡二叉树

【思路】

枚举每一个节点 判断该节点的左右子树高度是否相差1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public int getDepth(TreeNode root){
        
        if( root == null) return 0;
        
        int left = getDepth(root.left);
        int right = getDepth(root.right);
        
        return left > right ? left + 1 : right + 1;
        
    }
    
    public boolean isBalanced(TreeNode root) {
        if( root == null ) return true;
        
        //获得以root为根的左右子树的高度
        int left = getDepth(root.left);
        int right = getDepth( root.right);
        if( Math.abs(left - right) > 1 ) return false;
        
        return isBalanced(root.left) && isBalanced(root.right);
        
    }
}

标签：right,TreeNode,Offer,int,getDepth,二叉树,打卡,root,left
来源： https://blog.csdn.net/weixin_44855907/article/details/115269660