Binary Tree Level Order Traversal II

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],
]

Solution: Queue version. On the basis of 'Binary Tree Level Order Traversal', reverse the final vector.

　　　　　NULL is used to make gap between levels.

/**
 * 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) {
        vector<vector<int> > res;
        queue<TreeNode*> q;
        if(!root) return res;
        q.push(root);
        q.push(NULL);
        vector<int> level;
        
        while(true) {
            TreeNode* node = q.front();
            q.pop();
            if(node) {
                level.push_back(node->val);
                if(node->left) q.push(node->left);
                if(node->right) q.push(node->right);
            }
            else {
                res.push_back(level);
                level.clear();
                if(q.empty()) break;
                q.push(NULL);
            }
        }
        
        reverse(res.begin(), res.end());
        return res;
    }
};