平衡二叉树

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
LeetCode链接：https://leetcode-cn.com/problems/ping-heng-er-cha-shu-lcof/

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

        if not root:
            return True
        return abs(height(root.left) - height(root.right)) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)