# LC 199. Binary Tree Right Side View

### [Problem link](https://leetcode.com/problems/binary-tree-right-side-view/)

###### tags: `leedcode` `medium` `python` `c++` `Binary Tree`

Given the <code>root</code> of 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 1:**
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/14/tree.jpg" style="width: 401px; height: 301px;" />

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

**Example 2:**

Input: root = [1,null,3]
Output: [1,3]

**Example 3:**

Input: root = []
Output: []

**Constraints:**

- The number of nodes in the tree is in the range <code>[0, 100]</code>.
- <code>-100 <= Node.val <= 100</code>

## Solution 1
#### Python

# iteration
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        
        if not root:
            return []

        q = deque([root])

        while q:
            size_q = len(q)
            for i in range(size_q):
                node = q.popleft()
                if i == size_q - 1:
                    res.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)

        return res

# recursive
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        res = []

        def bfs(node, depth):
            if not node:
                return []
            if depth == len(res):
                res.append([])
            res[depth].append(node.val)
            if node.left:
                bfs(node.left, depth + 1)
            if node.right:
                bfs(node.right, depth + 1)

        bfs(root, 0)

        return [i[-1] for i in res]

#### C++

// iteration
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> res;
        queue<TreeNode*> q;

        if (root != nullptr) {
            q.push(root);
        }
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode *node = q.front();
                q.pop();
                if (i == size - 1) {
                    res.push_back(node->val);
                }
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
        }
        return res;
    }
};

// recursive
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> ans;

    void dfs(TreeNode *node, int depth) {
        if (node == nullptr) {
            return;
        }
        if (depth == ans.size()) {
            ans.push_back(node->val);
        }
        dfs(node->right, depth + 1);
        dfs(node->left, depth + 1);
    }

    vector<int> rightSideView(TreeNode* root) {
        dfs(root, 0);
        return ans;
    }
};

>### Complexity
>n = number of tree’s nodes
>l = number of tree's level nodes
>| | Time Complexity | Space Complexity |
>| ---------------------- | --------------- | ---------------- |
>| Solution 1(iteration) | O(n) | O(l) |
>| Solution 1(recursive) | O(n) | O(n) |

## Note
solution1:
iteration: 每次只處理一層, 所以SC: O(l)
recursive: 讀取完全部的node再輸出res, 所以SC: O(l)