LeetCode Populating Next Right Pointers in Each Node I and II

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题目：

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.

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

题意：

就是给定一棵任意的二叉树，然后调用这个connect函数后，就能够使得最后输出成，每一个节点的next指向同一层的后一个节点，每一层的最后一个节点的next为空，那么最后得到第二幅图的这种树的结构。此题就是典型的采用层次遍历来做。下面代码的方法不管是完全二叉树还是非完全二叉树，都是屡试不爽，都很成功，是一种非常好的层次遍历的方法。

public class Solution {    public void connect(TreeLinkNode root){LinkedList<TreeLinkNode> list = new LinkedList<TreeLinkNode>();if(root == null)return;else{list.add(root);int length = list.size();while(!list.isEmpty()){while(length-- > 0){TreeLinkNode node = list.peek();    list.poll();    if(length == 0)    node.next = null;    else     node.next = list.peek();    if(node.left != null)    list.add(node.left);    if(node.right != null)    list.add(node.right);}length = list.size();}}}}

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