Populating Next Right Pointers in Each Node --- leetcode

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).
一个满二叉树，每个节点添加指向右边节点的指针，没有右兄弟的就是NULL

思路：
广搜好像很容易实现，但是不能原地实现（需要队列）×
DFS肯定要用栈或者递归，O(lgn)空间 ×

有一个层的next指针已经确定好的话，其下一层的next指针很好确定，类似于广搜（直接用next实现了队列）
所以这题的做法;
从root开始，root的next = NULL
root的下一层依次指向右兄弟
……
完成。。。

代码如下：

/**
* 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)	return;
TreeLinkNode *rootTmp = root;
TreeLinkNode* QHead;
while (rootTmp->left)
{
QHead = rootTmp;
while (QHead)
{
QHead->left->next = QHead->right;
if (QHead->next)
QHead->right->next = QHead->next->left;
QHead = QHead->next;
}
rootTmp = rootTmp->left;
}
}
};

参考了大神代码，我是不太会做啦、、、