python判断平衡二叉树

题目：输入一棵二叉树，判断该二叉树是否是平衡二叉树。若左右子树深度差不超过1则为一颗平衡二叉树。
思路：

1. 使用获取二叉树深度的方法来获取左右子树的深度
2. 左右深度相减，若大于1返回False
3. 通过递归对每个节点进行判断，若全部均未返回False，则返回True

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 IsBalance_solution(self,pRoot):
        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.IsBalance_solution(pRoot.right) and self.IsBalance_solution(pRoot.left)