110. 平衡二叉树

https://leetcode-cn.com/problems/balanced-binary-tree/
思路：后序遍历，对root的左右子树递归，每次计算root的左右子树高，然后计算root的高，为左右较高者+1.若左右之差大于1，全局flag为false

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        self.flag = True
        self.help(root)
        return self.flag
        
    def help(self, root):
        if not root:
            return 0 
        
        lh = self.help(root.left)
        rh = self.help(root.right)
        height = lh if lh > rh else rh
        height += 1
        
        if abs(lh - rh) > 1:
            self.flag = False
            
        return height