How do you get the right-side view of a binary tree?

Q: How do you get the right-side view of a binary tree?

**Target Audience:** Beginners preparing for coding interviews — you should know what a binary tree is; BFS and DFS are explained here. **Difficulty:** Medium **Language:** Python 3 > **How to read this doc:** every idea is explained in plain words first. When a word appears that an interviewer will expect you to know, it is tagged `📌 Interview term:` with a one-line definition, and every tagged term is collected in the Quick Glossary at the end. Every program on this page was actually run, and the output you see in the **Verified** section is copied from that run. ## 1. Why This Even Matters — A Story First Imagine a tree-shaped stack of shelves standing in a dark room, each shelf holding one lamp. You walk around to the right-hand side and look straight at it. On every floor of the stack you see exactly one lamp: the one closest to you. Lamps further to the left on the same floor are hidden behind it — but if a floor has only one lamp, way over on the left, you still see that

The problem

Question presented to candidate: "Imagine standing on the right side of a binary tree. Write a Python function that returns, as a list, the values of the nodes you can see, ordered from top to bottom."

What a strong answer should cover:

The view is the rightmost node of each level — and it can come from the left subtree when the right side is shorter.
BFS: level-order traversal with a size snapshot, keeping the last node of each round.
DFS: visit node, right, left with a depth counter and record a node when depth equals the number of values so far.
Both are O(n) time; BFS uses O(w) space, DFS O(h).
Python details: collections.deque for O(1) popleft(), and view=None rather than a mutable default list in the DFS.
Why "follow the right pointers" is wrong, with a counterexample like [1, 2, 3, 4].

Clarifying questions expected:

One value per level, top to bottom, and an empty list for an empty tree?

Code / implementation expected: Yes — right_side_view(root) with a deque-based BFS, plus the DFS alternative, in Python 3.

trees
bfs

Try it yourself

Sketch your approach before you read the solution. That is what the interview tests.

Stuck? Get a hint.

A short nudge toward the approach. It will not give the answer away.

Solution

Target Audience: Beginners preparing for coding interviews — you should know what a binary tree is; BFS and DFS are explained here.
Difficulty: Medium
Language: Python 3

How to read this doc: every idea is explained in plain words first. When a word appears that an interviewer will expect you to know, it is tagged 📌 Interview term: with a one-line definition, and every tagged term is collected in the Quick Glossary at the end. Every program on this page was actually run, and the output you see in the Verified section is copied from that run.

1. Why This Even Matters — A Story First

Imagine a tree-shaped stack of shelves standing in a dark room, each shelf holding one lamp. You walk around to the right-hand side and look straight at it. On every floor of the stack you see exactly one lamp: the one closest to you. Lamps further to the left on the same flo

The solution is hidden so you can try first

11 min read · 11 sections

Python code

Edit the code and run it here. Each printed line with an expected value becomes a test.

PythonBFS: last node of every level (6 trees)

Practice

PythonBFS vs DFS, left view, and the right-pointer trap on 200 trees

Practice

Discussion

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