Maximum Width of Binary Tree in DSA Rust - Build from Scratch

Maximum Width of Binary Tree

📖 Scenario: You are working with a binary tree data structure that represents a family tree. You want to find the maximum width of this tree, which means the largest number of family members present at any single generation level.

🎯 Goal: Build a Rust program that creates a binary tree, sets up a queue for level order traversal, calculates the maximum width of the tree, and prints the result.

📋 What You'll Learn

Create a binary tree with the exact nodes and structure given

Use a queue to help traverse the tree level by level

Calculate the maximum width (maximum number of nodes at any level)

Print the maximum width as the final output

💡 Why This Matters

🌍 Real World

Finding the maximum width of a binary tree can help understand the broadest generation in family trees, organizational charts, or decision trees.

💼 Career

This concept is useful in software engineering roles that involve tree data structures, such as building parsers, UI component trees, or analyzing hierarchical data.

1

Create the Binary Tree Structure

Create a binary tree using the TreeNode struct with the exact structure: root node value 1, left child 2, right child 3, left child of 2 is 4, right child of 2 is 5, right child of 3 is 7. Use Option and Box as needed.

DSA Rust

// Your code here to create the binary tree

Hint

Use Rc and RefCell to allow multiple ownership and mutability of tree nodes.

2

Set Up a Queue for Level Order Traversal

Create a mutable VecDeque called queue and push the root node into it to prepare for level order traversal.

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;

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

impl TreeNode {
    pub fn new(val: i32) -> Rc> {
        Rc::new(RefCell::new(TreeNode { val, left: None, right: None }))
    }
}

fn main() {
    let root = TreeNode::new(1);
    root.borrow_mut().left = Some(TreeNode::new(2));
    root.borrow_mut().right = Some(TreeNode::new(3));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(4));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(5));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(7));

    // Create a mutable VecDeque called queue and push root into it
    // Your code here
}

Hint

Use VecDeque::new() to create the queue and push_back to add the root.

3

Calculate the Maximum Width of the Binary Tree

Write a while loop that runs while queue is not empty. Inside the loop, get the current level size with queue.len(). Update a variable max_width to the maximum of itself and the current level size. Then, for each node in the current level, pop it from the front of queue, and if it has a left child, push it to the back of queue. Similarly, if it has a right child, push it to the back of queue. Initialize max_width to 0 before the loop.

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;

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

impl TreeNode {
    pub fn new(val: i32) -> Rc> {
        Rc::new(RefCell::new(TreeNode { val, left: None, right: None }))
    }
}

fn main() {
    let root = TreeNode::new(1);
    root.borrow_mut().left = Some(TreeNode::new(2));
    root.borrow_mut().right = Some(TreeNode::new(3));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(4));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(5));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(7));

    let mut queue: VecDeque>> = VecDeque::new();
    queue.push_back(root);

    let mut max_width = 0;
    // Your code here to calculate max_width
}

Hint

Use a while loop with queue.len() to track level size. Update max_width if current level is wider.

4

Print the Maximum Width

Add a println! statement to print the value of max_width.

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;

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

impl TreeNode {
    pub fn new(val: i32) -> Rc> {
        Rc::new(RefCell::new(TreeNode { val, left: None, right: None }))
    }
}

fn main() {
    let root = TreeNode::new(1);
    root.borrow_mut().left = Some(TreeNode::new(2));
    root.borrow_mut().right = Some(TreeNode::new(3));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(4));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(5));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(7));

    let mut queue: VecDeque>> = VecDeque::new();
    queue.push_back(root);

    let mut max_width = 0;
    while !queue.is_empty() {
        let level_size = queue.len();
        if level_size > max_width {
            max_width = level_size;
        }
        for _ in 0..level_size {
            let node = queue.pop_front().unwrap();
            let node_borrow = node.borrow();
            if let Some(left) = &node_borrow.left {
                queue.push_back(Rc::clone(left));
            }
            if let Some(right) = &node_borrow.right {
                queue.push_back(Rc::clone(right));
            }
        }
    }

    // Print the max_width
    // Your code here
}

Hint

Use println!("{}", max_width); to print the maximum width.

Practice

1. What does the "maximum width" of a binary tree represent?

easy

A. The total number of nodes in the tree

B. The height of the tree from root to the deepest leaf

C. The largest number of nodes present at any single level of the tree

D. The sum of values of all nodes at the widest level

Solution

1. Step 1: Understand the definition of width in a binary tree

   The width of a binary tree at a level is the count of nodes present at that level.

2. Step 2: Identify maximum width

   The maximum width is the largest such count among all levels.

3. Final Answer:

   The largest number of nodes present at any single level of the tree -> Option C

4. Quick Check:

   Width = max nodes at any level [OK]

Hint: Width means count of nodes at widest level [OK]

Common Mistakes:

• Confusing width with tree height
• Thinking width is sum of node values
• Mixing width with total nodes count

2. Which Rust data structure is most suitable to track nodes level by level when calculating maximum width of a binary tree?

easy

A. VecDeque (double-ended queue)

B. Vec (vector)

C. HashMap

D. BTreeSet

Solution

1. Step 1: Identify the need for level order traversal

   To find maximum width, we visit nodes level by level, which is a breadth-first search (BFS).

2. Step 2: Choose data structure for BFS

   BFS uses a queue to process nodes in FIFO order. Rust's VecDeque supports efficient push/pop from both ends.

3. Final Answer:

   VecDeque (double-ended queue) -> Option A

4. Quick Check:

   Use queue (VecDeque) for BFS traversal [OK]

Hint: Use VecDeque for efficient queue operations in BFS [OK]

Common Mistakes:

• Using Vec which is inefficient for queue operations
• Using HashMap which is unordered
• Using BTreeSet which is for sorted unique elements

3. Given the following Rust code snippet for maximum width calculation, what is the printed output?

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

fn max_width(root: Option) -> usize {
    use std::collections::VecDeque;
    if root.is_none() { return 0; }
    let mut queue = VecDeque::new();
    queue.push_back(root.unwrap());
    let mut max_width = 0;
    while !queue.is_empty() {
        let level_size = queue.len();
        max_width = max_width.max(level_size);
        for _ in 0..level_size {
            let node = queue.pop_front().unwrap();
            if let Some(left) = node.left {
                queue.push_back(left);
            }
            if let Some(right) = node.right {
                queue.push_back(right);
            }
        }
    }
    max_width
}

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

medium

A. 3

B. 2

C. 4

D. 1

Solution

1. Step 1: Analyze the tree structure

   Level 1: Node 1 (1 node) Level 2: Nodes 2 and 3 (2 nodes) Level 3: Node 4 (1 node)

2. Step 2: Calculate maximum width

   Widths per level: 1, 2, 1 Maximum width is 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Max width = 2 nodes at level 2 [OK]

Hint: Count nodes per level, pick max count [OK]

Common Mistakes:

• Counting total nodes instead of per level
• Ignoring empty children in width calculation
• Confusing node values with count

4. The following Rust code attempts to compute the maximum width of a binary tree but has a bug. What is the bug?

fn max_width(root: Option) -> usize {
    use std::collections::VecDeque;
    if root.is_none() { return 0; }
    let mut queue = VecDeque::new();
    queue.push_back(root.unwrap());
    let mut max_width = 0;
    while !queue.is_empty() {
        let level_size = queue.len();
        max_width = max_width.max(level_size);
        for _ in 0..level_size {
            let node = queue.pop_front().unwrap();
            if node.left.is_some() {
                queue.push_back(node.left.unwrap());
            }
            if node.right.is_some() {
                queue.push_back(node.right.unwrap());
            }
        }
    }
    max_width
}

medium

A. Not handling empty tree case

B. Using VecDeque instead of Vec causes performance issues

C. Not updating max_width inside the loop

D. Calling unwrap() on Option without checking causes panic

Solution

1. Step 1: Identify unsafe unwrap calls

   Calling unwrap() on node.left or node.right without safely extracting causes panic if None.

2. Step 2: Correct approach

   Use if let Some(left) = node.left { queue.push_back(left); } to safely handle Options.

3. Final Answer:

   Calling unwrap() on Option without checking causes panic -> Option D

4. Quick Check:

   Unsafe unwrap causes runtime panic [OK]

Hint: Use if let to safely unwrap Options [OK]

Common Mistakes:

• Using unwrap() directly on Option fields
• Ignoring empty tree check (though present here)
• Confusing VecDeque with Vec performance

5. You want to find the maximum width of a binary tree where some levels have missing nodes between others (e.g., a sparse tree). Which approach correctly calculates the maximum width including gaps between nodes? A common method is to assign position indices to nodes as if the tree was a complete binary tree. Which Rust code snippet correctly updates the maximum width using position indices during level order traversal?

use std::collections::VecDeque;

fn max_width_with_positions(root: Option) -> usize {
    if root.is_none() { return 0; }
    let mut max_width = 0;
    let mut queue = VecDeque::new();
    queue.push_back((root.unwrap(), 0usize));
    while !queue.is_empty() {
        let level_length = queue.len();
        let first_index = queue.front().unwrap().1;
        let last_index;
        for i in 0..level_length {
            let (node, idx) = queue.pop_front().unwrap();
            let idx = idx - first_index; // normalize to prevent overflow
            if node.left.is_some() {
                queue.push_back((node.left.unwrap(), 2 * idx + 1));
            }
            if node.right.is_some() {
                queue.push_back((node.right.unwrap(), 2 * idx + 2));
            }
            if i == level_length - 1 {
                last_index = idx;
            }
        }
        max_width = max_width.max(last_index + 1);
    }
    max_width
}

hard

A. Normalize indices by subtracting first index at each level to avoid overflow

B. Use raw indices without normalization to get exact positions

C. Ignore indices and count nodes only for width

D. Use depth instead of indices for width calculation

Solution

1. Step 1: Understand position indexing

   Assigning indices as if complete binary tree helps measure width including gaps.

2. Step 2: Normalize indices to prevent overflow

   Subtracting first index at each level keeps indices small and prevents usize overflow.

3. Step 3: Calculate width as last_index - first_index + 1

   Using normalized indices, width = last_index + 1 at each level.

4. Final Answer:

   Normalize indices by subtracting first index at each level to avoid overflow -> Option A

5. Quick Check:

   Index normalization prevents overflow and calculates width correctly [OK]

Hint: Subtract first index each level to keep indices small [OK]

Common Mistakes:

• Not normalizing indices causing overflow
• Ignoring gaps by counting nodes only
• Using depth instead of position indices