# 199. Binary Tree Right Side View

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.

Example:

Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

# Solution

Either bfs or dfs, while making sure we keep the rightmost node -- need to visit all nodes to make sure we don't miss any children.

Approach 1: DFS.

Approach 2: BFS.

# Code (Python)

Approach 1:

from collections import deque
class Solution:
    def rightSideView(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        result = []
        queue = deque([(root, 0)])
        while queue:
            node, level = queue.popleft()
            if len(result) == level:
                result.append(node.val)
            else:
                result[-1] = node.val
            if node.left:
                queue.append((node.left, level + 1))
            if node.right:
                queue.append((node.right, level + 1))
        return result

# Code (C++)

Approach 1:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
// DFS.
class Solution {
private:
    vector<int> view;
    void doRightSideView(TreeNode* root, int level) {
        if (root == NULL)
            return;
        if (level == view.size())
            view.push_back(root->val);
        else
            view[level] = root->val;
        doRightSideView(root->left, level + 1);
        doRightSideView(root->right, level + 1);
    }
public:
    vector<int> rightSideView(TreeNode* root) {
        doRightSideView(root, 0);
        return view;
    }
};

Approach 2:

// BFS
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> view;
        queue<TreeNode*> q;
        if (root)
            q.push(root);
        while (!q.empty())
        {
            int qSize = q.size();
            TreeNode *node = NULL;
            for (int i = 0; i < qSize; ++i)
            {
                node = q.front();
                q.pop();
                if (node->left)
                    q.push(node->left);
                if (node->right)
                    q.push(node->right);
            }
            view.push_back(node->val);
        }
        return view;
    }
};