Populating Next Right Pointers in Each Node

[size=medium][/size] 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

注：本题用到了树的层序遍历，使用到了队列这种数据结构。

package leetcode;

import java.util.LinkedList;
import java.util.Queue;

class TreeLinkNode {
    int val;
    TreeLinkNode left, right, next;
    TreeLinkNode(int x) { val = x; }
}
public class PopulatingNextRightPointersInEachNode {
	public void connect(TreeLinkNode root) {
		if(root == null) return;//如果root为空，直接返回
        Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
        queue.offer(root);
        queue.offer(null);
        while(queue.size() > 1){//注意这里是1，队列中至少有两个元素，如果只有一个元素，说明已经执行到最后了
        	//得到队列中第一个元素，并出列
        	TreeLinkNode treeLinkNode = queue.poll();
        	if(treeLinkNode.left != null) queue.offer(treeLinkNode.left);
        	if(treeLinkNode.right != null)queue.offer(treeLinkNode.right);
        	//处理queue中的null
        	TreeLinkNode treeLinkNodeNext = queue.element();
        	if(treeLinkNodeNext == null) {
        		queue.poll();
        		queue.offer(null);
        	}
        	//处理next
        	treeLinkNode.next = treeLinkNodeNext;
        	//System.out.println((treeLinkNode.next == null) ? null : treeLinkNode.next.val);
        }
    }
	}