leetcode(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.

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

题目分析：平衡二叉树的判断，需要统计到左右子树的情况，看左右子树的高度差，所以，需要返回的信息是左右子树的高度的问题。

python代码实现：

# 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 isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root == None:
            return True
        left, right = self.height(root)
        if abs(left - right) > 1:
            return False
        if root.left:
            if self.isBalanced(root.left) == False:
                return False
        if root.right:
            if self.isBalanced(root.right) == False:
                return False
        return True

    def height(self, root):
        if root == None:
            return (0, 0)
        if root.left == None and root.right == None:
            return (0, 0)
        if root.left and root.right == None:
            return (1 + max(self.height(root.left)), 0)
        if root.right and root.left == None:
            return (0, 1 + max(self.height(root.right)))
        return (1 + max(self.height(root.left)), 1 + max(self.height(root.right)))

看一下大佬的程序：

class Solution(object):
    def isBalanced(self, root):

        def check(root):
            if root is None:
                return 0
            left  = check(root.left)
            right = check(root.right)
            if left == -1 or right == -1 or abs(left - right) > 1:
                return -1
            return 1 + max(left, right)

        return check(root) != -1

程序里递归，还是第一次见呀，好好学习一下这种写法。

class Solution(object):
    def isBalanced(self, root):
        stack, node, last, depths = [], root, None, {}
        while stack or node:
            if node:
                stack.append(node)
                node = node.left
            else:
                node = stack[-1]
                if not node.right or last == node.right:
                    node = stack.pop()
                    left, right  = depths.get(node.left, 0), depths.get(node.right, 0)
                    if abs(left - right) > 1: return False
                    depths[node] = 1 + max(left, right)
                    last = node
                    node = None
                else:
                    node = node.right
        return True

这个是采用迭代的方法来写的，也是晚上需要好好看一下的。