110. 平衡二叉树

https://leetcode-cn.com/problems/balanced-binary-tree/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-09-14 17:25:33
 * @LastEditTime: 2020-09-14 17:27:19
 * @Description: https://leetcode-cn.com/problems/balanced-binary-tree/
 * @FilePath: \leetcode-googtech\#110. Balanced Binary Tree\Solution.java
 * @WebSite: https://algorithm.show/
 */

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

class Solution {

    boolean result = true;

    // BFS : 深度优先遍历
    public boolean isBalanced(TreeNode root) {
        dfs(root);
        return result;
    }
    private int dfs(TreeNode root) {
        if(root == null) return 0;
        int lDepth = dfs(root.left);
        int rDepth = dfs(root.right);
        if(Math.abs(lDepth - rDepth) > 1) result = false;
        return Math.max(lDepth, rDepth) + 1;
    }
}

Python

'''
Author: Goog Tech
Date: 2020-09-14 17:25:59
LastEditTime: 2020-09-14 17:28:45
Description: https://leetcode-cn.com/problems/balanced-binary-tree/
FilePath: \leetcode-googtech\#110. Balanced Binary Tree\Solution.py
WebSite: https://algorithm.show/
'''

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

# 递归的三个条件 : 左平衡,右平衡,当前节点平衡.
class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        if not root: return True
        return self.isBalanced(root.left) and self.isBalanced(root.right) and abs(self.height(root.left) - self.height(root.right)) <= 1

    def height(self,root:TreeNode) -> int:
        if not root: return 0
        return max(self.height(root.left), self.height(root.right)) + 1