Coding Interview PatternsBinary Tree Right Side View
MediumTree Depth-First Search

Binary Tree Right Side View

Explanation & Solution

Practice Problem

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.

Examples

Example 1:

```tree

[1, 2, 3, null, 5, null, 4]

`

Input:root = [1, 2, 3, null, 5, null, 4]
0
1
1
2
2
3
3
null
4
5
5
null
6
4
Output:[1, 3, 4]
0
1
1
3
2
4

Explanation: From the right side, you see node 1 at level 0, node 3 at level 1, and node 4 at level 2.

Example 2:

```tree

[1, null, 3]

`

Input:root = [1, null, 3]
0
1
1
null
2
3
Output:[1, 3]
0
1
1
3

Example 3:

Input: root = []

Output: []

Constraints

  • The number of nodes in the tree is in the range [0, 100]
  • -100 <= Node.val <= 100

Approach

Tree Depth-First Search pattern

1. Handle Empty Tree

  • If root is null:
  • Return []

2. Initialize BFS

  • Create result = [] for the right-side view values
  • Create queue = [root] with head = 0 for efficient dequeue

3. Process Level by Level

  • While the queue is not empty:
  • Calculate levelSize = queue.length - head
  • This tells us how many nodes are on the current level

4. Find the Rightmost Node

  • Loop through all nodes in the current level:
  • Dequeue via queue[head++]
  • If this is the last node in the level (i === levelSize - 1):
  • Push its value to result
  • Enqueue left and right children

5. Why the Last Node?

  • BFS processes nodes left to right
  • The last node at each level is the rightmost node
  • That is the node visible from the right side

6. Return Result

  • After all levels, return result
  • Contains one value per level — the rightmost node's value

Key Insight

  • Standard BFS level-by-level traversal
  • Only keep the last node from each level
  • A left-side-only subtree can still appear in the right side view if the right subtree is shorter
Time
O(n)visit each node once
Space
O(n)queue holds at most one level of nodes

Visualization

Input:
DFS Traversal[1, 2, 3, 5, 4]
12354
output0

Press play to start DFS traversal

CurrentIn stackComplete
11 steps

Solution Code

Previous
Build Binary Tree from Preorder and Inorder Traversal
Next
Lowest Common Ancestor of a Binary Tree