原题地址：https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/

题意：

1       /  \      2    3     / \  / \    4  5  6  7 变为：

1 -> NULL       /  \      2 -> 3 -> NULL     / \  / \    4->5->6->7 -> NULL

解题思路：看到二叉树我们就想到需要使用递归的思路了。直接贴代码吧，思路不难。

代码：

# Definition for a  binary tree node# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = None#         self.next = Noneclass Solution:    # @param root, a tree node    # @return nothing    def connect(self, root):        if root and root.left:            root.left.next = root.right            if root.next:                root.right.next = root.next.left            else:                root.right.next = None            self.connect(root.left)            self.connect(root.right)