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---
title: '199. Binary Tree Right Side View'
description: Given the root 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
icon: dot
topics:
- { name: "Tree", slug: "tree" }
- { name: "Depth-First Search", slug: "depth-first-search" }
- { name: "Breadth-First Search", slug: "breadth-first-search" }
- { name: "Binary Tree", slug: "binary-tree" }
video: "d4zLyf32e3I"
sidebar:
label: 'Binary Tree Right Side View'
badge: 'Medium'
---
### Example 1:
- Input: `root = [1,2,3,null,5,null,4]`
- Output: `[1,3,4]`
### Example 2:
- Input: `root = [1,2,3,4,null,null,null,5]`
- Output: `[1,3,4,5]`
### Example 3:
- Input: `root = [1,null,3]`
- Output: `[1,3]`
### Example 4:
- Input: `root = []`
- Output: `[]`
### Constraints:
- The number of nodes in the tree is in the range [0, 100].
- `-100 <= Node.val <= 100`
## Approach
```mermaid
flowchart TD
S(["rightSideView(root)"]) --> I["res = [], q = deque([root])"]
I --> W{"q non-empty?"}
W -- no --> E(["return res"])
W -- yes --> L["rightSide = None, qLen = len(q) — qLen freezes the current level's size"]
L --> F{"more i in range(qLen)?"}
F -- no --> V{"rightSide is not None?"}
V -- yes --> A["res.append(rightSide.val) — the last real node popped on this level is the rightmost"]
A --> W
V -- no --> W
F -- yes --> P["node = q.popleft()"]
P --> N{"node is not None?"}
N -- no --> F
N -- yes --> C["rightSide = node, q.append(node.left), q.append(node.right)"]
C --> F
```
## Solution
```py
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
from collections import deque
class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
res = []
q = deque([root])
while q:
rightSide = None
qLen = len(q)
for i in range(qLen):
node = q.popleft()
if node:
rightSide = node
q.append(node.left)
q.append(node.right)
if rightSide:
res.append(rightSide.val)
return res
```