LeetCode Binary Tree Right Side View

Description

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,
1 <—
/
2 3 <—
\
5 4 <—
You should return [1, 3, 4].

The original problem is here.

The original code is here.

My Solution

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

/*Binary Tree Right Side View
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<stack>
#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<int> rightSideView(TreeNode* root) {
        vector<int> value;
        if(root == NULL)
            return value;
        
        vector<TreeNode*> oneLevel;
        vector<vector<TreeNode*>> myTree;
        TreeNode *node;

        oneLevel.push_back(root);
        myTree.push_back(oneLevel);
        while(oneLevel.size()>=1){
            vector<TreeNode*> nextLevel;
            for(int i=0;i<oneLevel.size();i++)
            {
                node = oneLevel[i];
                if(node->left != NULL)
                    nextLevel.push_back(node->left);
                if(node->right != NULL)
                    nextLevel.push_back(node->right);
            }
            if(nextLevel.size()>=1)
                myTree.push_back(nextLevel);
            oneLevel = nextLevel;
        }
        for(int i=0;i<myTree.size();i++){
            vector<TreeNode*> level = myTree[i];
            int size = level.size();
            node = level[size-1];
            value.push_back(node->val);
        }
        return value;
    }
};

Note

To solve the problem, we use breadth-first search to search the tree. We get the all values of final node in one level.