Binary Tree Right Side View

Question

Accepted Answer

To solve the Binary Tree Right Side View problem, I would start by clarifying that we need the rightmost node at every depth level. Given the requirement to return values top-to-bottom, a Level Order Traversal using Breadth-First Search is the most straightforward approach. I would initialize a queue and add the root node. While the queue is not empty, I would determine the number of nodes currently in the queue, which represents the size of the current level. I would then iterate exactly that many times. Inside this loop, I would dequeue a node and check if it is the last node in the current level. If it is, I add its value to my result list. For non-last nodes, I simply enqueue their children if they exist. This ensures we only capture the rightmost visible node per level. Alternatively, a recursive DFS could work by visiting the right child before the left, keeping track of the current depth and adding the node value only when the depth matches the size of our result list, effectively filling gaps from top to bottom. Both approaches run in O(N) time since we visit every node once. The BFS approach uses O(W) space for the queue, while DFS uses O(H) space for the recursion stack. I prefer BFS here for its clarity in handling level boundaries, which aligns well with systems requiring predictable memory usage patterns.

Binary Tree Right Side View

Why Interviewers Ask This

How to Answer This Question

Key Points to Cover

Sample Answer

Common Mistakes to Avoid

Sound confident on this question in 5 minutes

Related Interview Questions

How do you implement a queue using two stacks?

Find K Closest Elements (Heaps)

Convert Binary Tree to Doubly Linked List in Place