平衡二叉树Python解法

给定一个二叉树，判断它是否是高度平衡的二叉树。

本题中，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 。

例：

输入：root = [3,9,20,null,null,15,7]
输出：true

解析：

写一个深度函数，计算左子树和右子树的深度，然后进行比对，对子树也进行判断。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):def depth(self, root):  # 计算深度函数if root is None:return 0else:return 1 + max(self.depth(root.left), self.depth(root.right))def isBalanced(self, root):""":type root: TreeNode:rtype: bool"""if root is None:return Trueif abs(self.depth(root.left) - self.depth(root.right)) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right):  # 判断是否为平衡树return Trueelse:return False

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