110. Balanced Binary Tree [Easy] 平衡树/递归

110. Balanced Binary Tree

110. Balanced Binary Tree

啊啊啊，今天再做这道题竟然不会了，要设置一个辅助函数，这个函数返回值是根结点的深度，但是如果左子树或者右子树不是平衡树，就返回-1，主函数只要不是-1，就是True

# 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
        """
        return self.helper(root) != -1
    def helper(self, root):
        if root == None:
            return 0
        left = self.helper(root.left)
        if left == -1:
            return -1
        right = self.helper(root.right)
        if right == -1:
            return -1
        if abs(left-right) > 1:
            return -1
        return 1 + max(left, right)