110. Balanced Binary Tree

problem

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.

判断是否平衡二叉树，（相邻子节点深度差值不超过1）

solution

• 父节点要获取子节点状态，在二叉树里面是个逆向的流程
• 如果先比较子节点，然后比较父节点，如何实现？压栈？

discuss

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

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