Right Side View of Binary Tree in DSA Rust - Execution Trace

Concept Flow - Right Side View of Binary Tree

Start at root node

↓

Initialize queue with root

↓

While queue not empty

↓

Get number of nodes at current level

↓

For each node in level

↓

Pop node

↓

If last node in level

↓

Record node value for right view

↓

Repeat for next level

↓

End when queue empty

↓

Return collected right side view values

We traverse the tree level by level from left to right, recording the last node's value at each level to get the right side view.

Execution Sample

DSA Rust

use std::collections::VecDeque;

struct TreeNode {
    val: i32,
    left: Option<Box<TreeNode>>,
    right: Option<Box<TreeNode>>,
}

fn right_side_view(root: Option<Box<TreeNode>>) -> Vec<i32> {
    let mut result = vec![];
    let mut queue = VecDeque::new();
    if let Some(node) = root {
        queue.push_back(node);
    }
    while !queue.is_empty() {
        let level_size = queue.len();
        for i in 0..level_size {
            let node = queue.pop_front().unwrap();
            if i == level_size - 1 {
                result.push(node.val);
            }
            if let Some(left) = node.left {
                queue.push_back(left);
            }
            if let Some(right) = node.right {
                queue.push_back(right);
            }
        }
    }
    result
}

This code collects the rightmost node's value at each level of the binary tree using a queue for level order traversal.

Execution Table

Step	Operation	Node Visited	Queue State	Right Side View Collected	Visual Tree State
1	Initialize queue with root	1	[1]	[]	1 / \ 2 3 / \ 4 5
2	Process level 1 (size=1)	1	[]	[1]	1 / \ 2 3 / \ 4 5
3	Add children of 1	1	[2, 3]	[1]	1 / \ 2 3 / \ 4 5
4	Process level 2 (size=2)	2	[3]	[1]	1 / \ 2 3 / \ 4 5
5	Add children of 2	2	[3, 4]	[1]	1 / \ 2 3 / \ 4 5
6	Process level 2 (size=2)	3	[4]	[1, 3]	1 / \ 2 3 / \ 4 5
7	Add children of 3	3	[4, 5]	[1, 3]	1 / \ 2 3 / \ 4 5
8	Process level 3 (size=2)	4	[5]	[1, 3]	1 / \ 2 3 / \ 4 5
9	Add children of 4	4	[5]	[1, 3]	1 / \ 2 3 / \ 4 5
10	Process level 3 (size=2)	5	[]	[1, 3, 5]	1 / \ 2 3 / \ 4 5
11	Add children of 5	5	[]	[1, 3, 5]	1 / \ 2 3 / \ 4 5
12	Queue empty, end traversal	-	[]	[1, 3, 5]	1 / \ 2 3 / \ 4 5

💡 Queue is empty, all levels processed, right side view collected

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 7	After Step 10	Final
queue	[]	[1]	[2, 3]	[4, 5]	[]	[]
result	[]	[]	[1]	[1, 3]	[1, 3, 5]	[1, 3, 5]
level_size	0	1	2	2	2	0

Key Moments - 3 Insights

Why do we record the node value only when i == level_size - 1?

Because the last node at each level is the rightmost node visible from the right side. This is shown in execution_table rows 2, 6, and 10 where the last node's value is added to the result.

Why do we add left child before right child to the queue if we want the right side view?

We add the left child before the right child to ensure that nodes at each level are enqueued and thus processed from left to right. This makes the rightmost node the last one processed per level. Reversing the order would process right to left, recording leftmost nodes instead (see vq3).

What happens if the tree is empty (root is None)?

The queue remains empty and the while loop never runs, so the result stays empty. This is implied by the initial check in the code and no steps occur in the execution_table for an empty tree.

Visual Quiz - 3 Questions

Look at the execution_table at Step 6, what is the queue state?

A [4]

B [3]

C [4, 5]

D [2, 3]

💡 Hint

Check the 'Queue State' column at Step 6 in the execution_table.

At which step is the value 5 added to the right side view?

A Step 8

B Step 10

C Step 6

D Step 12

💡 Hint

Look at the 'Right Side View Collected' column in the execution_table to see when 5 appears.

If we changed the code to add right child before left child to the queue, how would the right side view change?

A It would remain the same [1, 3, 5]

B It would be [1, 2, 4]

C It would be empty

D It would be [1, 3, 4]

💡 Hint

Changing the enqueue order to right before left reverses the processing order within each level (right-to-left), so the last node per level becomes the leftmost visible from right, yielding [1, 2, 4] for this tree.

Concept Snapshot

Right Side View of Binary Tree:
- Traverse tree level by level (BFS).
- Use a queue to hold nodes of current level.
- For each level, record the last node's value.
- Add children left then right to queue.
- Result is list of rightmost nodes per level.

Full Transcript

This visualization shows how to find the right side view of a binary tree by traversing it level by level using a queue. We start with the root node in the queue. For each level, we process all nodes, adding their children to the queue. We record the value of the last node processed at each level because it is the rightmost visible node from that side (assuming left-to-right processing order). The execution table tracks each step, showing the node visited, queue state, and the right side view collected so far. The variable tracker shows how the queue and result list change after key steps. Key moments clarify why we record only the last node per level and why the order of adding left before right child is crucial for correct left-to-right level processing. The quiz tests understanding of queue states and result collection at specific steps. This method ensures we see the tree as if looking from the right side, capturing the visible nodes in order.

Practice

1. What does the "Right Side View of Binary Tree" represent?

easy

A. Nodes visible when the tree is viewed from the right side

B. Nodes visible when the tree is viewed from the left side

C. All leaf nodes of the binary tree

D. All nodes at the deepest level of the tree

Solution

1. Step 1: Understand the view perspective

   The right side view shows nodes visible when looking at the tree from the right side only.

2. Step 2: Differentiate from other views

   Unlike left side or leaf views, it only includes the rightmost nodes at each level.

3. Final Answer:

   Nodes visible when the tree is viewed from the right side -> Option A

4. Quick Check:

   Right side view = rightmost nodes [OK]

Hint: Right side view shows rightmost nodes at each level [OK]

Common Mistakes:

• Confusing right side view with left side view
• Including all leaf nodes instead of rightmost nodes
• Thinking it shows all nodes at the deepest level

2. Which Rust code snippet correctly declares a binary tree node struct for the right side view problem?

easy

A. struct TreeNode { val: i32, left: &TreeNode, right: &TreeNode }

B. struct TreeNode { val: i32, left: Option, right: Option }

C. struct TreeNode { val: i32, left: TreeNode, right: TreeNode }

D. struct TreeNode { val: i32, left: Option<TreeNode>, right: Option<TreeNode> }

Solution

1. Step 1: Check ownership and optional children

   Rust needs Option for nullable child nodes with ownership.

2. Step 2: Validate correct struct syntax

   Option allows recursive tree structure safely.

3. Final Answer:

   struct TreeNode { val: i32, left: Option, right: Option } -> Option B

4. Quick Check:

   Use Option for children [OK]

Hint: Use Option for nullable child nodes [OK]

Common Mistakes:

• Using direct TreeNode type causing infinite size error
• Using references without lifetimes
• Using Option<TreeNode> without Box causing compile error

3. What is the output of the right side view function for this tree?

    1
   / \
  2   3
   \   \
    5   4

Assuming the function returns a vector of node values visible from the right side.

medium

A. [1, 2, 4]

B. [1, 2, 5]

C. [1, 3, 4]

D. [1, 3, 5]

Solution

1. Step 1: Identify rightmost nodes at each level

   Level 0: 1, Level 1: 3 (right child), Level 2: 4 (right child of 3)

2. Step 2: Collect these nodes in order

   Right side view vector is [1, 3, 4]

3. Final Answer:

   [1, 3, 4] -> Option C

4. Quick Check:

   Rightmost nodes per level = [1, 3, 4] [OK]

Hint: Pick rightmost node at each tree level [OK]

Common Mistakes:

• Choosing left child nodes instead of rightmost
• Including nodes not visible from right side
• Mixing order of nodes in output

4. Given this Rust function snippet for right side view, what is the bug?

fn right_side_view(root: Option) -> Vec<i32> {
    let mut result = Vec::new();
    let mut queue = Vec::new();
    if root.is_some() {
        queue.push(root.unwrap());
    }
    while !queue.is_empty() {
        let level_size = queue.len();
        for i in 0..level_size {
            let node = queue.remove(0);
            if i == level_size - 1 {
                result.push(node.val);
            }
            if node.left.is_some() {
                queue.push(node.left.unwrap());
            }
            if node.right.is_some() {
                queue.push(node.right.unwrap());
            }
        }
    }
    result
}

medium

A. The function returns the left side view, not right side

B. Queue should be a linked list, not a vector

C. The loop should iterate backwards

D. Using unwrap() consumes the Option, causing ownership errors

Solution

1. Step 1: Analyze use of unwrap() on Option

   Calling unwrap() moves the value out, so subsequent uses cause ownership errors.

2. Step 2: Identify correct approach

   Use references or cloning to avoid moving ownership when pushing children into queue.

3. Final Answer:

   Using unwrap() consumes the Option, causing ownership errors -> Option D

4. Quick Check:

   unwrap() moves ownership, causing errors [OK]

Hint: Avoid unwrap() that moves ownership inside loops [OK]

Common Mistakes:

• Ignoring ownership rules in Rust
• Thinking vector is wrong data structure here
• Misunderstanding loop direction relevance

5. How would you modify a standard level order traversal in Rust to return the right side view of a binary tree efficiently?

hard

A. Traverse each level left to right, record last node's value per level

B. Traverse each level right to left, record first node's value per level

C. Traverse each level left to right, record first node's value per level

D. Traverse each level right to left, record last node's value per level

Solution

1. Step 1: Understand level order traversal direction

   Standard level order goes left to right per level.

2. Step 2: Identify which node to record

   Recording the last node at each level gives the rightmost visible node.

3. Final Answer:

   Traverse each level left to right, record last node's value per level -> Option A

4. Quick Check:

   Last node per level in left-to-right traversal = right side view [OK]

Hint: Record last node at each level in left-to-right traversal [OK]

Common Mistakes:

• Recording first node instead of last
• Changing traversal direction unnecessarily
• Confusing left and right side views