LeetCode 110. Balanced Binary Tree

0. 题目

Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:

a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

一棵高度平衡二叉树定义为：一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过1。

1. c++版本

#include 
class Solution {
public:
    int heightOfTree(TreeNode* root) {
        if (!root)
            return 0;
        return max(heightOfTree(root->left), heightOfTree(root->right)) + 1;
    }
    bool isBalanced(TreeNode* root) {
        if (!root)
            return true;
        if (abs(heightOfTree(root->left) - heightOfTree(root->right)) > 1)
            return false;
        return isBalanced(root->left) && isBalanced(root->right);
    }
};

2. python版本

class Solution(object):
    def heightOfTree(self, root):
        if not root:
            return 0
        return max(self.heightOfTree(root.left), self.heightOfTree(root.right)) + 1

    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        if abs(self.heightOfTree(root.left) - self.heightOfTree(root.right)) > 1:
            return False
        return self.isBalanced(root.left) and self.isBalanced(root.right)

版本2. https://leetcode.com/problems/balanced-binary-tree/discuss/35708/VERY-SIMPLE-Python-solutions-(iterative-and-recursive)-both-beat-90

class Solution(object):
    def isBalanced(self, root):
            
        def check(root):
            if root is None:
                return 0
            left  = check(root.left)
            right = check(root.right)
            if left == -1 or right == -1 or abs(left - right) > 1:
                return -1
            return 1 + max(left, right)
            
        return check(root) != -1