Leetcode Populating Next Right Pointers in Each Node

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.
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

分析：

这道题的实质上是分层遍历，有两个地方需要注意。

1. root的next一定是null

2. 在分层遍历的时候，最本层的前n-1个节点处理方法一样，next是下一个节点，但是最后一个节点的next是null

Java代码实现：

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
if(root==null)
return;

root.next = null;
Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
if(root.left!=null)
q.offer(root.left);
if(root.right!=null)
q.offer(root.right);

while(!q.isEmpty())
{
int size = q.size();
for(int i=0;i<size;i++)
{
TreeLinkNode node = q.poll();
if(i!=size-1)
node.next = q.peek();
else
node.next = null;
if(node.left!=null)
q.offer(node.left);
if(node.right!=null)
q.offer(node.right);
}
}
}
}