Given a binary tree

    struct TreeLinkNode {      TreeLinkNode *left;      TreeLinkNode *right;      TreeLinkNode *next;    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1       /        2    3     / \  /     4  5  6  7

After calling your function, the tree should look like:

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

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

思路：关键是5跟6如何连接，用递归的做法，把当前root的左节点和它所有后继右节点 与 右节点和所有深度的左节点连接。

public class Solution {    public void connect(TreeLinkNode root) {        if(root==null)            return;        if(root.left!=null){            TreeLinkNode l = root.left;            TreeLinkNode r = root.right;            while(l!=null){                l.next=r;                l=l.right;                r=r.left;            }                    }        connect(root.left);        connect(root.right);                    }}

参考：

http://blog.csdn.net/havenoidea/article/details/12840497

http://blog.csdn.net/fightforyourdream/article/details/14514165