Populating Next Right Pointers in Each Node

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 <br/>
   /  \ <br/>
  2    3 <br/>
  

  / \ / \
  4 5 6 7
  After calling your function, the tree should look like:
  

     1 -> NULL <br/>
   /  \   <br/>
  2 -> 3 -> NUL <br/>
  

  / \ / \
  4->5->6->7 -> NULL

题目大意：添加next指针指向二叉树中节点的右边

题目难度：Medium

import java.util.*;

/**
 * Created by gzdaijie on 16/6/9
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        Queue<TreeLinkNode> queue = new LinkedList<>();
        if (root == null) return;

        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            TreeLinkNode pre = null, p;
            while (size-- > 0) {
                p = queue.poll();
                if (pre != null) pre.next = p;
                pre = p;
                if (p.left != null) queue.offer(p.left);
                if (p.right != null) queue.offer(p.right);
            }
        }
    }
}