107 Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

Solution:

/**
 * Definition for a binary tree node.
 * 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) {
        vector<vector<int>> res;
        if ( root == NULL ) return res;
        stack<vector<TreeNode*>> record;
        vector<TreeNode*> res1;
        res1.push_back(root);
        while ( res1.size() != 0 ) {
            record.push(res1);
            int nres = res1.size();
            vector<TreeNode*> res2;
            for ( int i = 0; i <= nres-1; i++ ) {
                if ( res1[i]->left != NULL ) res2.push_back(res1[i]->left);
                if ( res1[i]->right != NULL ) res2.push_back(res1[i]->right);
            }
            res1 = res2;
        }
        while ( ! record.empty() ) {
            res1 = record.top();
            record.pop();
            int nres = res1.size();
            vector<int> tmp;
            for ( int i = 0; i <= nres-1; i++ ) tmp.push_back(res1[i]->val);
            res.push_back(tmp);
        }
        return res;
    }
};

Notes
DETAILS!