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],
]

====================================================================

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > result;
        if(NULL == root){
            return result;
        }
        vector<TreeNode *> v1;
        vector<TreeNode *> v2;
        vector<int> line;
        v1.push_back(root);
        while(!v1.empty()){
            line.clear();
            for (int i = 0; i < v1.size(); i++){
                line.push_back(v1[i]->val);
                if(v1[i]->left) v2.push_back(v1[i]->left);
                if(v1[i]->right) v2.push_back(v1[i]->right);
            }
            result.push_back(line);
            v1 = v2;
            v2.clear();
        }
            
        //reverse result vector using a temp stack
        stack<vector<int> >s;
        for (int i = 0; i < result.size(); i++){
            s.push(result[i]);
        }
        result.clear();
        while(!s.empty()){
            result.push_back(s.top());
            s.pop();
        }
        return result;
    }
};

Just revers result vetor using a temp stack, nothing else, AC in one submission.