Right Side View of Binary Tree
A medium Binary Trees problem included in Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Binary Trees
- Sheets
- 2
- Core for
- 4 roles
- Platform
- LeetCode
The problem
Given the root of a binary tree, return the values of the nodes visible from the right side, ordered from top to bottom. Only the rightmost node at each depth level is visible.
Example 1
- Input
- [1,2,3,null,5,null,4]
- Output
- [1,3,4]
- Why
- Rightmost nodes at levels 0, 1, and 2 are 1, 3, and 4 respectively.
Example 2
- Input
- [1,null,3]
- Output
- [1,3]
- Why
- Level 0 has 1, level 1 has 3 as the rightmost node.
Example 3
- Input
- [1,2,3,4,null,null,5]
- Output
- [1,3,5]
- Why
- Rightmost at level 0: 1, level 1: 3, level 2: 5.
Constraints
- The number of nodes is in the range [1, 100].
- -100 <= Node.val <= 100
How to think about it
Updated 2026-09-09The right view needs exactly one node per horizontal level: the last one seen from left to right. Either traverse BFS and pluck the final element of each level's batch, or traverse DFS visiting right subtrees before left subtrees, capturing the very first node that reaches each new depth.
Approaches, worst first
BFS level-order extraction
time O(n) · space O(w)
Perform standard queue-based level order traversal. In each level loop, capture the value of the last node popped from the queue and push it into the answer. Very intuitive but stores whole levels in the queue.
DFS right-first recursionWrite this one
time O(n) · space O(h)
Recurse with `(node, depth)`. If `depth === result.length`, this is the first time visiting this depth, so append `node.val`. Recurse into `node.right` first, then `node.left`. Directly appends rightmost nodes with zero auxiliary arrays.
Where people lose marks · 3
- Traversing only right child pointers (`curr = curr.right`); if the right subtree is shorter than the left, deeper nodes on the left branch become visible from the right.
- In DFS right-first, recursing left before right appends the left view instead.
- Failing to account for an empty tree returns `[null]` or throws if root is null.
The theory behind it
Binary Trees — the ground this problem stands on. All Binary Trees problems
What Binary Trees is
A binary tree is a branching data structure that starts at a single top node called the root, like an upside-down family tree. Every node holds a piece of data and can branch out to at most two children below it, known as the left child and the right child. Because there is no ordering rule about which values go left or right, finding a specific item can require checking every single node in the entire tree.
When to reach for it
Reach for binary trees when problems present hierarchical data with left and right child pointers. Questions asking for tree height, maximum depth, path sums from root to leaf, diameter, lowest common ancestor, or checking whether two trees are mirror reflections of each other all signal binary tree traversals. Any problem asking to inspect or reconstruct a tree layer by layer or path by path belongs here.
How the pattern works
Think recursively by focusing on what a single node must do. If the current node is null, return the base answer immediately. Otherwise, ask the left child for its result, ask the right child for its result, and combine both answers with the current node value before returning up to the parent. For horizontal scans, use a queue to read nodes layer by layer, measuring the queue length at the start of each layer to group nodes by depth.
What each operation costs
| Operation | Time |
|---|---|
| traverse all nodes using recursion or queue | O(n) |
| search for an arbitrary value in an unordered tree | O(n) |
| call stack memory on balanced tree | O(log n) |
| call stack memory on skewed tree | O(n) |
What usually goes wrong with Binary Trees
- Dereferencing left or right child pointers without checking if the current node is null, throwing null pointer errors on empty trees or leaf nodes.
- Defining a leaf node incorrectly by stopping when either child is null instead of checking that both left and right children are simultaneously null.
- Computing tree diameter by taking left height plus right height inside a recursive helper without updating a global maximum across every visited node.
Which roles need this problem
Binary Trees is a core topic for these 4 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Full-Stack Developer, Android Developer, iOS Developer.
Track this in your role's order
Pick your target role and all 370 problems — including this one — resequence to what that interview actually asks. Free.
Start freeMore Binary Trees problems
Problem set and role mapping as of .