Populating Next Right Pointers in Each Node 更好的方法待补充

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

第一感觉是用层序遍历，虽然ac了，但是空间复杂度似乎不满足常数的要求。

/**
* 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 == 0)
return;

queue<TreeLinkNode *> q;
q.push(root);
q.push(0);

while(!q.empty())
{
TreeLinkNode * cur = q.front();
q.pop();

if(cur)
{
if(cur -> left)
q.push(cur -> left);
if(cur -> right)
q.push(cur -> right);
if(!q.empty())
if(q.front() == 0)
cur -> next = NULL;
else
cur -> next = q.front();
}
else
if(!q.empty())
q.push(0);

}

}
};

待补充