[leetcode: Python] 110. Balanced Binary Tree

题目：
Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

题意：
判断给定的二叉树是否是平衡树。

平衡二叉树（Balanced Binary Tree）又被称为AVL树（有别于AVL算法），且具有以下性质：它是一 棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树。平衡二叉树的常用实现方法有红黑树、AVL、替罪羊树、Treap、伸展树等。

方法一：性能115ms

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

class Solution(object):
    def height(self, root):
        if root == None:
            return True
        return max(self.height(root.left), self.height(root.right)) + 1

    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root == None:
            return True
        elif abs(self.height(root.left) - self.height(root.right)) <= 1:
            return self.isBalanced(root.left) and self.isBalanced(root.right)
        else:
            return False

方法二：性能62ms

class Solution(object):
    def isBalanced(self, root):
        global res
        res = True
        def getHeight(root):
            global res
            if not root: return 0
            leftHeight = getHeight(root.left)
            if leftHeight < 0:
                res = False
                return -1
            rightHeight = getHeight(root.right)
            if rightHeight < 0 or abs(leftHeight - rightHeight) > 1:
                res = False
                return -1
            return max(leftHeight, rightHeight) + 1
        getHeight(root)
        return res

第二种方法再好好看看！