Right Side View of Binary Tree in DSA Rust - Build from Scratch

Right Side View of Binary Tree

📖 Scenario: Imagine you have a tree of family members, and you want to see only the members visible from the right side. This means you want to see the rightmost member at each level of the family tree.

🎯 Goal: You will build a program that shows the right side view of a binary tree. The program will print the values of the nodes visible when looking at the tree from the right side.

📋 What You'll Learn

Create a binary tree with specific nodes

Use a variable to track the current level during traversal

Implement a function to find the right side view of the tree

Print the list of node values visible from the right side

💡 Why This Matters

🌍 Real World

Right side view of a binary tree helps in visualizing hierarchical data from a specific perspective, useful in UI layouts, organizational charts, and decision trees.

💼 Career

Understanding tree traversal and view extraction is important for software engineers working with data structures, algorithms, and system design.

1

Create the Binary Tree Structure

Create a binary tree with the following nodes and structure using Option and Box in Rust:
root node with value 1,
left child with value 2,
right child with value 3,
left child of node 2 with value 4,
right child of node 3 with value 5.

DSA Rust

// Define the TreeNode struct and create the binary tree as described
// Your code here

Hint

Use Option> for left and right children to represent possible absence of nodes.

2

Add a Level Tracker Variable

Add a mutable variable called current_level and set it to 0. This will help track the current depth level during tree traversal.

DSA Rust

#[derive(Debug)]
struct TreeNode {
    val: i32,
    left: Option>,
    right: Option>,
}

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode {
            val: 2,
            left: Some(Box::new(TreeNode {
                val: 4,
                left: None,
                right: None,
            })),
            right: None,
        })),
        right: Some(Box::new(TreeNode {
            val: 3,
            left: None,
            right: Some(Box::new(TreeNode {
                val: 5,
                left: None,
                right: None,
            })),
        })),
    }));

    // Create a mutable variable current_level and set it to 0
    // Your code here
}

Hint

Use let mut current_level = 0; inside main function.

3

Implement Right Side View Function

Write a function called right_side_view that takes a reference to an Option> and a mutable reference to current_level, and returns a Vec containing the values visible from the right side. Traverse the tree prioritizing the right child first, then the left child, and add the node's value to the result vector only if the current level is equal to the length of the result vector.

DSA Rust

#[derive(Debug)]
struct TreeNode {
    val: i32,
    left: Option>,
    right: Option>,
}

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode {
            val: 2,
            left: Some(Box::new(TreeNode {
                val: 4,
                left: None,
                right: None,
            })),
            right: None,
        })),
        right: Some(Box::new(TreeNode {
            val: 3,
            left: None,
            right: Some(Box::new(TreeNode {
                val: 5,
                left: None,
                right: None,
            })),
        })),
    }));

    let mut current_level = 0;

    // Define the right_side_view function below
    // Your code here
}

Hint

Use recursion to visit the right child first, then the left child. Add the node's value to result only if level == result.len().

4

Print the Right Side View Result

Print the result vector that contains the right side view node values using println!.

DSA Rust

#[derive(Debug)]
struct TreeNode {
    val: i32,
    left: Option>,
    right: Option>,
}

fn right_side_view(node: &Option>, level: usize, result: &mut Vec) {
    if let Some(n) = node {
        if level == result.len() {
            result.push(n.val);
        }
        right_side_view(&n.right, level + 1, result);
        right_side_view(&n.left, level + 1, result);
    }
}

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode {
            val: 2,
            left: Some(Box::new(TreeNode {
                val: 4,
                left: None,
                right: None,
            })),
            right: None,
        })),
        right: Some(Box::new(TreeNode {
            val: 3,
            left: None,
            right: Some(Box::new(TreeNode {
                val: 5,
                left: None,
                right: None,
            })),
        })),
    }));

    let mut current_level = 0;
    let mut result = Vec::new();
    right_side_view(&root, current_level, &mut result);

    // Print the result vector
    // Your code here
}

Hint

Use println!("{:?}", result); to print the vector of visible node values.

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