Populating Next Right Pointers in Each NodeII

题目描述：

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 toNULL.

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

这个题层次遍历来做，和前面的Binary Tree Zigzag Level Order Traversal差不多，注意最后一行要拿出来计算一遍就行了。

代码如下：

public class Solution {
    class Node {
        int level;
        TreeLinkNode treeLinkNode;
        Node(TreeLinkNode treeLinkNode,int level) { 
            this.treeLinkNode=treeLinkNode;
            this.level=level;
        }
    }
    public void connect(TreeLinkNode root) {
        Deque<Node> queue=new ArrayDeque<Node>();
        if(root==null)
            return;
        queue.offer(new Node(root, 1));
        int curlevel=0;
        List<TreeLinkNode> list=new ArrayList<TreeLinkNode>();
        while(!queue.isEmpty()){
            Node node=queue.poll();
            if(node.level>curlevel){
                if(!list.isEmpty()){
                    for(int i=0;i<list.size()-1;i++){
                        list.get(i).next=list.get(i+1);
                    }
                }
                curlevel=node.level;
                list.clear();
            }
            list.add(node.treeLinkNode);
            if(node.treeLinkNode.left!=null)
                queue.offer(new Node(node.treeLinkNode.left, node.level+1));
            if(node.treeLinkNode.right!=null)
                queue.offer(new Node(node.treeLinkNode.right, node.level+1));
        }
        for(int i=0;i<list.size()-1;i++){
            list.get(i).next=list.get(i+1);
        }
    }
}