题解 | #判断是不是平衡二叉树#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param pRoot TreeNode类 
# @return bool布尔型
#
class Solution:
    def IsBalanced_Solution(self , pRoot: TreeNode) -> bool:
        # write code here
        if pRoot is None:
            return True
        
        h_left = self.get_height(pRoot.left)
        h_right = self.get_height(pRoot.right)

        if (abs(h_left-h_right)<=1): #得保证所有左右子树都是平衡二叉树
            return self.IsBalanced_Solution(pRoot.right) and self.IsBalanced_Solution(pRoot.left)
        return False

    def get_height(self, pRoot: TreeNode) -> int:# 计算树的深度
        
        if pRoot is None:
            return 0
        if pRoot.left is None and pRoot.right is None:
            return 1

        return max(self.get_height(pRoot.left), self.get_height(pRoot.right))+1

这道题考察“平衡二叉树”和“树的深度的计算”。