--- 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" issue: "https://github.com/prdlk/leetcode/issues/107" 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 ```