leetcode-117. Populating Next Right Pointers in Each Node II

题目：

Follow up for problem “Populating Next Right Pointers in Each Node”.

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.
For example,
Given the following binary tree,
1
/ \
2 3
/ \ \
4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL

和上一题不同。这里结构不是固定的了所以需要使用队列去保存当前的信息。而且有很多需要注意的小细节。

首先从思路上来说还是用FIFO的队列，并且每个循环体内都替换一次。直到到最后一行队列为空则终止循环。
外层循环每一层循环异常。内层循环针对当前对立的每个节点。

/** * 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 ;        LinkedList<TreeLinkNode> list = new LinkedList<TreeLinkNode>();        list.add(root);        while(list.size()!=0){            LinkedList<TreeLinkNode> replace = new LinkedList<TreeLinkNode>();            TreeLinkNode pre = list.remove();            if(pre.left!=null) replace.add(pre.left);            if(pre.right!=null) replace.add(pre.right);            TreeLinkNode cur = null;            while(list.size()!=0){                cur = list.remove();                if(cur!=null && cur.left!=null) replace.add(cur.left);                if(cur!=null && cur.right!=null) replace.add(cur.right);                pre.next = cur;                if(cur!=null) pre = cur;            }            if(replace.size()!=0) replace.add(null);            list = replace;        }    }}