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],
] 

Solution: the basic idea is similar to Binary Tree Level Order Traversal I. [Another good solution: 1]

/**
 * 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
        if(root==NULL){
            vector<vector<int> > ret;
            return ret;
        }
        vector<vector<int> > left, right;
        left = levelOrderBottom(root->left);
        right = levelOrderBottom(root->right);
        reverse(left.begin(), left.end());
        reverse(right.begin(), right.end());
        int numLayer = min(left.size(), right.size());
        for(int i=0; i<numLayer; i++){
            left[i].insert(left[i].end(), right[i].begin(), right[i].end());
        }
        if(left.size()<right.size()){
            left.insert(left.end(), right.begin()+numLayer, right.end());
        }
        left.insert(left.begin(), vector<int>(1, root->val));
        reverse(left.begin(), left.end());
        return left;
    }
};