LeetCode Binary Tree Level Order Traversal2

Description

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]
]

The original problem is here.

The original code is here.

My Solution

I solve this problem in C++, as below:

/*
*Binary Tree Level Order Traversal
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<stack>
#include<vector>
#include<stdlib.h>
using namespace std;

/**
 * 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<int> vec;
        vector<vector<int>> res;
        TreeNode* node;
        vector<TreeNode*> currentLevel, nextLevel;
        if(root==NULL)
            return res;
        if(root!=NULL)
            currentLevel.push_back(root);
        do{
            for(int i=0;i<currentLevel.size();i++){
                node = currentLevel[i];
                vec.push_back(node->val);
                if(node->left!=NULL)
                    nextLevel.push_back(node->left);
                if(node->right!=NULL)
                    nextLevel.push_back(node->right);
            }
            res.push_back(vec);
            vec.clear();
            currentLevel.clear();
            currentLevel = nextLevel;
            nextLevel.clear();
        }while(!currentLevel.empty());
        
        int size = res.size()-1;
        for(int i=0;i<=size/2;i++){
            vec = res[i];
            res[i] = res[size-i];
            res[size-i] = vec;
        }
        return res;
    }
};

Note

The problem is almost the same to the “Binary Tree Level Order Traversal”, just reverse the final result.