# 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.