110. Balanced Binary Tree

原文链接：

110. Balanced Binary Tree

读题：

判断是否为平衡二叉树，平衡二叉树的定义是每个节点的左右子树的高度差不超过1，显然需要通过求高度，也就是树的深度差来判断。

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def maxDepth(self, root):        if not root:            return 0        return max(self.maxDepth(root.left),self.maxDepth(root.right)) + 1     def isBalanced(self, root):        """        :type root: TreeNode        :rtype: bool        """        if not root:            return True        lengthL = self.maxDepth(root.left)        lengthR = self.maxDepth(root.right)        if lengthL > lengthR + 1 or lengthR > lengthL + 1:            return False        else:            return self.isBalanced(root.left) and self.isBalanced(root.right)