Boundary Traversal of Binary Tree in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a binary tree node struct with left and right children.

DSA Rust

struct Node {
    data: i32,
    left: Option<Box<Node>>,
    right: Option<Box<Node>>,
}

impl Node {
    fn new(data: i32) -> Self {
        Node {
            data,
            left: [1],
            right: None,
        }
    }
}

Drag options to blanks, or click blank then click option'

ANone

BSome(Box::new(Node::new(data)))

CSome(Node::new(data))

DBox::new(Node::new(data))

Attempts:

3 left

💡 Hint

Common Mistakes

Using Some(...) instead of None for initial children

Trying to assign a Node directly without Box or Option

The left child is initially empty, so we use None to indicate no child.

2fill in blank

medium

Complete the code to check if a node is a leaf (no children).

DSA Rust

fn is_leaf(node: &Option<Box<Node>>) -> bool {
    if let Some(n) = node {
        n.left.is_none() && n.right.[1]()
    } else {
        false
    }
}

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Ais_some

Bclone

Cunwrap

Dis_none

Attempts:

3 left

💡 Hint

Common Mistakes

Using is_some() which checks for presence, not absence

Trying to unwrap without checking

To check if the right child is empty, use is_none().

3fill in blank

hard

Fix the error in the function that prints the left boundary excluding leaves.

DSA Rust

fn print_left_boundary(node: &Option<Box<Node>>) {
    if let Some(n) = node {
        if !is_leaf(node) {
            print!("{} ", n.data);
            if n.left.is_some() {
                print_left_boundary(&n.[1]);
            } else {
                print_left_boundary(&n.right);
            }
        }
    }
}

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Aleft

Bright

Cdata

Dclone

Attempts:

3 left

💡 Hint

Common Mistakes

Recursing on right child instead of left

Trying to access data instead of child nodes

To continue down the left boundary, recurse on n.left.

4fill in blank

hard

Fill both blanks to collect leaf nodes in left-to-right order.

DSA Rust

fn print_leaves(node: &Option<Box<Node>>) {
    if let Some(n) = node {
        print_leaves(&n.[1]);
        if is_leaf(node) {
            print!("{} ", n.data);
        }
        print_leaves(&n.[2]);
    }
}

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Aleft

Bright

Cdata

Dclone

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right children in recursion

Printing node data before checking if leaf

To print leaves left to right, recurse left child first, then right child.

5fill in blank

hard

Fill all three blanks to print the boundary traversal of the binary tree.

DSA Rust

fn print_boundary(root: &Option<Box<Node>>) {
    if root.is_none() {
        return;
    }
    let r = root.as_ref().unwrap();
    print!("{} ", r.data); // root
    print_left_boundary(&r.[1]);
    print_leaves(&r.[2]);
    print_leaves(&r.[3]);
    print_right_boundary(&r.right);
}

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Aleft

Bright

Attempts:

3 left

💡 Hint

Common Mistakes

Using right child for left boundary

Not printing leaves from both sides

Start left boundary from left child, leaves from left and right children respectively.

Practice

1. What is the main purpose of the Boundary Traversal of Binary Tree?

easy

A. To print nodes in post-order traversal

B. To print nodes level by level

C. To print nodes in in-order traversal

D. To print all nodes on the boundary in anti-clockwise direction

Solution

1. Step 1: Understand boundary traversal definition

   Boundary traversal prints nodes on the edges of the tree: left boundary, leaves, and right boundary.

2. Step 2: Identify traversal direction

   The traversal is done in anti-clockwise order starting from root, left boundary, leaves, then right boundary.

3. Final Answer:

   To print all nodes on the boundary in anti-clockwise direction -> Option D

4. Quick Check:

   Boundary traversal = anti-clockwise boundary nodes [OK]

Hint: Boundary traversal prints edges anti-clockwise [OK]

Common Mistakes:

• Confusing boundary traversal with level order traversal
• Including internal nodes not on boundary
• Printing right boundary in clockwise order

2. Which of the following Rust function signatures is correct for a function that performs boundary traversal on a binary tree node defined as struct Node { val: i32, left: Option, right: Option }?

easy

A. fn boundary_traversal(root: Node) -> Vec<i32>

B. fn boundary_traversal(root: &Option) -> Vec<i32>

C. fn boundary_traversal(root: &Node) -> Vec<i32>

D. fn boundary_traversal(root: Option<Node>) -> Vec<i32>

Solution

1. Step 1: Analyze ownership and borrowing in Rust

   Binary tree nodes are wrapped in Option to allow null and heap allocation. Passing by reference avoids ownership transfer.

2. Step 2: Match function signature with correct parameter type

   Using &Option allows borrowing the root node safely without moving ownership.

3. Final Answer:

   fn boundary_traversal(root: &Option) -> Vec<i32> -> Option B

4. Quick Check:

   Borrow Option for traversal [OK]

Hint: Use &Option to borrow tree root safely [OK]

Common Mistakes:

• Passing Node by value causes ownership move
• Using Option<Node> instead of Option
• Using &Node without Option wrapper

3. Given the following Rust code snippet for boundary traversal, what will be the output printed for the tree below? Tree structure: ``` 1 / \ 2 3 / \ \ 4 5 6 / \ / 7 8 9 ``` Code snippet:

fn main() {
    let root = Some(Box::new(Node {
        val: 1,
        left: Some(Box::new(Node {
            val: 2,
            left: Some(Box::new(Node { val: 4, left: None, right: None })),
            right: Some(Box::new(Node {
                val: 5,
                left: Some(Box::new(Node { val: 7, left: None, right: None })),
                right: Some(Box::new(Node { val: 8, left: None, right: None })),
            })),
        })),
        right: Some(Box::new(Node {
            val: 3,
            left: None,
            right: Some(Box::new(Node {
                val: 6,
                left: Some(Box::new(Node { val: 9, left: None, right: None })),
                right: None,
            })),
        })),
    }));
    let result = boundary_traversal(&root);
    println!("{:?}", result);
}

medium

A. [1, 2, 4, 7, 8, 9, 6, 3]

B. [1, 2, 4, 7, 8, 9, 6, 3, 5]

C. [1, 2, 4, 7, 8, 5, 9, 6, 3]

D. [1, 2, 4, 5, 7, 8, 9, 6, 3]

Solution

1. Step 1: Identify boundary nodes in anti-clockwise order

   Root: 1; Left boundary (excluding leaves): 2; Leaves (left-to-right): 4, 7, 8, 9; Right boundary (excluding leaves, bottom-up): 6, 3.

2. Step 2: Combine nodes without duplicates

   Boundary traversal = [1, 2, 4, 7, 8, 9, 6, 3]

3. Final Answer:

   [1, 2, 4, 7, 8, 9, 6, 3] -> Option A

4. Quick Check:

   Boundary nodes = [1,2,4,7,8,9,6,3] [OK]

Hint: List left boundary, leaves, then right boundary reversed [OK]

Common Mistakes:

• Including internal nodes like 5 in boundary
• Adding right boundary top-down instead of bottom-up
• Missing leaf nodes on right subtree

4. The following Rust function attempts to perform boundary traversal but has a bug. What is the main issue?

fn print_left_boundary(node: &Option) {
    if let Some(n) = node {
        if n.left.is_some() {
            println!("{}", n.val);
            print_left_boundary(&n.left);
        } else if n.right.is_some() {
            println!("{}", n.val);
            print_left_boundary(&n.right);
        }
    }
}

medium

A. It incorrectly recurses on the right child when the left child is absent

B. It does not print the root node

C. It causes infinite recursion on leaf nodes

D. It skips nodes that have both left and right children

Solution

1. Step 1: Analyze recursion logic

   The function recurses on the left child if present, but falls back to recursing on the right child if no left child. For left boundary, always prefer and recurse only on left child (stop if absent).

2. Step 2: Understand boundary traversal rule for left boundary

   This incorrect fallback includes right subtree nodes in the left boundary traversal.

3. Final Answer:

   It incorrectly recurses on the right child when the left child is absent -> Option A

4. Quick Check:

   Left boundary: recurse left only [OK]

Hint: Always recurse on left child only for left boundary [OK]

Common Mistakes:

• Recursing on right child when left is absent in left boundary
• Not handling None nodes properly
• Incorrectly printing leaves in left boundary

5. You want to modify the boundary traversal to print nodes in clockwise order starting from root. Which of the following changes is correct to achieve this?

hard

A. Print leaves first, then left boundary, then right boundary, then root last

B. Print root, then left boundary top-down, leaves left to right, then right boundary bottom-up

C. Print root, then right boundary top-down, leaves right to left, then left boundary bottom-up

D. Print right boundary bottom-up, leaves left to right, then left boundary top-down, root last

Solution

1. Step 1: Understand clockwise boundary traversal order

   Clockwise traversal starts at root, then right boundary top-down, leaves from right to left, then left boundary bottom-up.

2. Step 2: Match the option with correct order

   Print root, then right boundary top-down, leaves right to left, then left boundary bottom-up matches this order exactly, ensuring correct clockwise traversal.

3. Final Answer:

   Print root, then right boundary top-down, leaves right to left, then left boundary bottom-up -> Option C

4. Quick Check:

   Clockwise boundary = root + right + leaves + left reversed [OK]

Hint: Clockwise = root, right boundary down, leaves right-left, left boundary up [OK]

Common Mistakes:

• Using anti-clockwise order for clockwise traversal
• Printing leaves left to right instead of right to left
• Reversing boundaries incorrectly