leetcode@ [116/117] Populating Next Right Pointers in Each Node I & II (Tree, BFS)

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

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

/**
* 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 node {
public:
TreeLinkNode* nd;
int lv;
node(TreeLinkNode* rhs, int l): nd(rhs), lv(l) {}
};

class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root)  return;

queue<node> q;
q.push(node(root, 0));
vector<node> vec;

while(!q.empty()) {
node top = q.front();
vec.push_back(top);
q.pop();

int cur_lv = top.lv;
TreeLinkNode* cur_nd = top.nd;

if(cur_nd->left)  q.push(node(cur_nd->left, cur_lv+1));
if(cur_nd->right)  q.push(node(cur_nd->right, cur_lv+1));
}

int l = 0, r = 1;
while(r < vec.size()) {
vec[l].nd->next = NULL;
while(r < vec.size() && vec[r].lv == vec[l].lv) {
vec[l].nd->next = vec[r].nd;
++l;
++r;
}

l = r;
r = l+1;
}
vec[vec.size()-1].nd->next = NULL;
}
};