给定一个二叉树，判断它是否是平衡二叉树。

• 二叉平衡树的左右子树高度相差不超过1，且左右子树均是二叉平衡树
• 二叉平衡树是所有结点的左右子树高度差均不超过1的二叉树

主要考对二叉平衡树的概念

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def isBalanced(self, root: Optional[TreeNode]) -> bool:# 二叉平衡树的左右子树高度相差不超过1，且左右子树均是二叉平衡树# 二叉平衡树是所有结点的左右子树高度差均不超过1的二叉树return self.judge(root)[0]def judge(self, root):if not root:return True, 0left_judge, left_height = True, 0if root.left:left_judge, left_height = self.judge(root.left)right_judge, right_height = True, 0if root.right:right_judge, right_height = self.judge(root.right)return (left_judge and right_judge and abs(left_height - right_height) < 2,max(left_height, right_height) + 1,)