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).

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

 Summary: The simplest way is BFS, but we need non-constant extra space. So, I traverses the tree Pre-order, and uses one node pointer for every level.

     void traverse(TreeLinkNode *root, int depth, vector<TreeLinkNode *> &current) {
         if(root->left != NULL && root->right != NULL){
             if(current.size() < depth + )
                 current.push_back(root->right);
             else{
                 current[depth]->next = root->left;
                 current[depth] = root->right;
             }
             root->left->next = root->right;
             //traverse the left subtree
             traverse(root->left, depth + , current);
             //traverse the right subtree
             traverse(root->right, depth + , current);
         }
     }

     void connect(TreeLinkNode *root) {
         if(root == NULL)
             return;

         vector<TreeLinkNode *> current;
         traverse(root, , current);
     }