"Leetcode" Binary Tree level Order traversal

A daily practice,

Given a binary tree, return the level order traversal of its nodes ' values. (ie, from left-to-right, level by level).

For example:
Given binary Tree {3,9,20,#,#,15,7} ,

3/9 20/15 7

Return its level order traversal as:

[3], [9,20], [15,7]]

The code is as follows:

Class solution {public:    int level (Treenode *node)      {        if  (node == null)  return 0;         if  (node->left == null &&  Node->right == null)  return 1;        return  max (Level (Node->left), Level (node->right))  + 1;    }     void levelpush (treenode *node, int height, vector<int> & Vec, int cur_height) {        if  (height ==  Cur_height)  vec.push_back (node->val);        if  (node- >left != null)  levelpush (node->left, height, vec, cur_height + 1 ); &NBSP;&NBSP;&NBSP;&NBSP;&Nbsp;   if  (node->right != null)  levelpush (Node->right, height, &NBSP;VEC,&NBSP;CUR_HEIGHT&NBSP;+&NBSP;1);     }    vector<vector <int> > levelorder (treenode *root)  {         int height = level (root);         vector<vector <int> > vvec;        for (int i = 1;i  <= height;i++)         {             vector<int> vec;             levelpush (root, i, vec, 1);             vvec.push_back (VEC);         }        return vvec;    }}; 

"Leetcode" Binary Tree level Order traversal