題目：輸入一棵二叉樹，判斷該二叉樹是否是平衡二叉樹。若左右子樹深度差不超過1則為一顆平衡二叉樹。
思路：1、使用獲取二叉樹深度的方法來獲取左右子樹的深度
2、左右深度相減，若大於1返回False
3、通過遞迴對每個節點進行判斷，若全部均未返回False，則返回True

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution
 
:
    def getDeepth(self, Root):
        if Root is None:
            return 0
        nright = self.getDeepth(Root.right)
        nleft = self.getDeepth(Root.left)
        return max(nright, nleft)+1
    def IsBalanced_Solution(self, pRoot):
        # write code here
        if pRoot is None:
            return
 
 True
        right = self.getDeepth(pRoot.right)
        left = self.getDeepth(pRoot.left)
        if abs(right - left) >1:
            return False
        return self.IsBalanced_Solution(pRoot.right) and self.IsBalanced_Solution(pRoot.left)

進階思路：
採用後續遍歷，當遍歷到根節點時，每個節點只會遍歷一次
以下程式無法執行，僅提供思路

class
 
 Solution:
    def IsBalanced(self, pRoot, depth):
        if pRoot is None:
            return True
        if self.IsBalanced(pRoot.right, right) and self.IsBalanced(pRoot.left, left):
            diff = abs(left - right)
            if diff <= 1:
                depth = max(left, right) +1
                return True
        return False
    def IsBalanced_Solution(self, pRoot):
        # write code here
        depth = 0
        return self.IsBalanced(pRoot, depth)