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 = None class 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)
[leetcode]Populating Next Right Pointers in Each Node @ Python,布布扣,bubuko.com
[leetcode]Populating Next Right Pointers in Each Node @ Python
原文：http://www.cnblogs.com/zuoyuan/p/3745170.html