Binary Tree Level Order Traversal II [LeetCode]

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

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

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

BFS solution:

     vector<vector<int> > levelOrderBottom(TreeNode *root) {
         vector<vector<int> > ret;
         if(root == NULL)
             return ret;
         vector<TreeNode *> level;
         level.push_back(root);
         while(true) {
             if(level.size() == )
                 break;
             vector<int> nums;
             vector<TreeNode *> tmp;
             for(auto item : level) {
                 nums.push_back(item->val);
                 if(item->left != NULL)
                     tmp.push_back(item->left);
                 if(item->right != NULL)
                     tmp.push_back(item->right);
             }
             ret.insert(ret.begin(), nums);
             level = tmp;
         }
         return ret;
     }

DFS solution:

     void getLevelNums(TreeNode *root, vector<vector<int> > &ret, int level) {
         if(ret.size() < level + ){
             vector<int> nums;
             nums.push_back(root->val);
             ret.insert(ret.begin(), nums);
         }else if (ret.size() >= level + ) {
             ret[ret.size() - level - ].push_back(root->val);
         }

         if(root->left != NULL)
             getLevelNums(root->left, ret, level + );

         if(root->right != NULL)
             getLevelNums(root->right, ret, level + );

     }
     vector<vector<int> > levelOrderBottom(TreeNode *root) {
         vector<vector<int> > ret;
         if(root == NULL)
             return ret;
         getLevelNums(root, ret, );
         return ret;
     }