LeetCode Populating Next Right Pointers in Each Node I and II

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();
}
}
}
}