Why Boundary Traversal of Binary Tree in DSA Rust?

The Big Idea

What if you could trace the outline of a complex tree without missing a single edge node or getting lost inside?

The Scenario

Imagine you have a family tree drawn on paper, and you want to trace the outer edge of the tree to see all the relatives standing on the boundary. Doing this by hand means checking every branch and leaf carefully, which can be confusing and slow.

The Problem

Manually tracing the boundary of a tree is slow because you must remember which nodes are on the left edge, which are leaves, and which are on the right edge. It's easy to miss nodes or count some twice, especially when the tree is large or uneven.

The Solution

Boundary traversal automatically walks around the tree's edges in order: left boundary, leaves, then right boundary. This method ensures every boundary node is visited once, making the process clear, fast, and error-free.

Before vs After

✗ Before

fn manual_boundary(nodes: &Vec<i32>) {
    // Check each node manually if it is on boundary
    for node in nodes {
        if is_left_or_right_boundary(node) || is_leaf(node) {
            print!("{} ", node);
        }
    }
}

✓ After

fn boundary_traversal(root: &Option<Box<TreeNode>>) {
    print_left_boundary(root);
    print_leaves(root);
    print_right_boundary(root);
}

What It Enables

This traversal lets you quickly see the outline of a tree, useful for visualizing structure or solving problems that need edge information.

Real Life Example

In a city map represented as a tree, boundary traversal helps find all the streets on the city's outer edge, useful for planning routes or security patrols.

Key Takeaways

Manual boundary checks are slow and error-prone.

Boundary traversal visits left edge, leaves, then right edge in order.

This method ensures all boundary nodes are covered once, clearly and efficiently.

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