LeetCode—110. Balanced Binary Tree

LeetCode—110. Balanced Binary Tree

题目

https://leetcode.com/problems/balanced-binary-tree/description/
判断一棵二叉树是否是平衡二叉树。
平衡二叉树指的是，两棵子树的任一节点的深度相差不超过1.

思路及解法

首先写一个函数用来得到每一个子树的深度，然后递归调用原函数，判断两棵子树的深度差是否超过1

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root==null) return true;
        int leftDepth = getDepth(root.left);
        int rightDepth = getDepth(root.right);
        if(Math.abs(leftDepth-rightDepth)>1) return false;
        else return isBalanced(root.left)&&isBalanced(root.right);
    }
    public int getDepth(TreeNode node){
        if(node==null) return 0;
        else if(node.left==null && node.right==null) return 1;
        else{
            int leftDepth = getDepth(node.left);
            int rightDepth = getDepth(node.right);
            return 1+(leftDepth>rightDepth?leftDepth:rightDepth);
        }
    }
}