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##（一）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
##（二）解题
参考： 【一天一道LeetCode】#116. Populating Next Right Pointers in Each Node
没想到昨天做的解法直接把今天的题给解了。一模一样的代码，请跳转到上题的解题思路：
/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(root==NULL) return;
queue<TreeLinkNode *> myque;
myque.push(root);
while(!myque.empty())
{
queue<TreeLinkNode *> temp_que;
TreeLinkNode *pre = NULL;
while(!myque.empty())
{
TreeLinkNode *temp = myque.front();
myque.pop();
if(pre==NULL) pre = temp;//相当于每一层的头节点
else {
pre->next = temp;//用next指针连接每一层的节点
pre = temp;
}
if(temp->left!=NULL) temp_que.push(temp->left);//下一层
if(temp->right!=NULL) temp_que.push(temp->right);//下一层
}
pre->next=NULL;//最后一个节点的next需要指向NULL
myque=temp_que;//进入下一层操作
}
}
};