Maximum Width of Binary Tree in DSA Rust

Mental Model

We want to find the widest level in a tree by counting nodes including gaps between them.

Analogy: Imagine a tree as a family photo where some people stand far apart; width counts all spaces between the first and last person in each row.

      1
     / \
    2   3
   /     \
  4       5

Levels:
0: 1
1: 2 -> 3
2: 4 -> null -> null -> 5

Dry Run Walkthrough

Input: Binary tree: 1 / \ 2 3 / \ 4 5

Goal: Find the maximum width of the tree, counting null gaps between nodes at each level.

Step 1: Start BFS at root with index 0

[ (1,0) ]

Why: We assign index 0 to root to track positions for width calculation

Step 2: Process level 0: nodes [(1,0)]

Width = 0 - 0 + 1 = 1

Why: Only one node, width is 1

Step 3: Add children of 1: (2,1), (3,2)

[ (2,1), (3,2) ]

Why: Left child index = 2*0+1=1, right child index = 2*0+2=2

Step 4: Process level 1: nodes [(2,1), (3,2)]

Width = 2 - 1 + 1 = 2

Why: Nodes at indices 1 and 2, width is 2

Step 5: Add children: 2 has left child (4,3), 3 has right child (5,6)

[ (4,3), (5,6) ]

Why: Left child index = 2*1+1=3, right child index = 2*2+2=6

Step 6: Process level 2: nodes [(4,3), (5,6)]

Width = 6 - 3 + 1 = 4

Why: Indices 3 and 6 with gaps count as width 4

Result:

Maximum width = 4

Annotated Code

DSA Rust

use std::collections::VecDeque;

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

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

fn width_of_binary_tree(root: Option<Box<TreeNode>>) -> usize {
    if root.is_none() {
        return 0;
    }
    let mut max_width = 0;
    let mut queue: VecDeque<(Box<TreeNode>, usize)> = VecDeque::new();
    queue.push_back((root.unwrap(), 0));

    while !queue.is_empty() {
        let level_length = queue.len();
        let first_index = queue.front().unwrap().1;
        let mut last_index = first_index;

        for _ in 0..level_length {
            let (node, index) = queue.pop_front().unwrap();
            last_index = index;

            if let Some(left) = node.left {
                queue.push_back((left, 2 * (index - first_index) + 1));
            }
            if let Some(right) = node.right {
                queue.push_back((right, 2 * (index - first_index) + 2));
            }
        }
        let width = last_index - first_index + 1;
        if width > max_width {
            max_width = width;
        }
    }
    max_width
}

fn main() {
    // Construct the tree:
    //       1
    //      / \
    //     2   3
    //    /     \
    //   4       5
    let mut root = Box::new(TreeNode::new(1));
    let mut node2 = Box::new(TreeNode::new(2));
    let mut node3 = Box::new(TreeNode::new(3));
    let node4 = Box::new(TreeNode::new(4));
    let node5 = Box::new(TreeNode::new(5));

    node2.left = Some(node4);
    node3.right = Some(node5);
    root.left = Some(node2);
    root.right = Some(node3);

    let max_width = width_of_binary_tree(Some(root));
    println!("Maximum width = {}", max_width);
}

queue.push_back((root.unwrap(), 0));

Initialize queue with root node at index 0

let first_index = queue.front().unwrap().1;

Record first node index at current level to normalize indices

for _ in 0..level_length {

Process all nodes at current level

let (node, index) = queue.pop_front().unwrap();

Pop node and its index from queue

last_index = index;

Update last node index at this level

queue.push_back((left, 2 * (index - first_index) + 1));

Add left child with normalized index

queue.push_back((right, 2 * (index - first_index) + 2));

Add right child with normalized index

let width = last_index - first_index + 1;

Calculate width of current level

if width > max_width { max_width = width; }

Update maximum width if current level is wider

Output

Maximum width = 4

Complexity Analysis

Time: O(n) because we visit each node once in BFS

Space: O(n) because queue can hold up to all nodes at the widest level

vs Alternative: Naive approach might store all nodes including nulls explicitly, increasing space; this method uses index math to track gaps efficiently

Edge Cases

Empty tree

Returns width 0 immediately

DSA Rust

if root.is_none() { return 0; }

Tree with single node

Width is 1 as only root exists

DSA Rust

queue.push_back((root.unwrap(), 0));

Tree with missing children creating gaps

Width counts gaps correctly using indices

DSA Rust

queue.push_back((left, 2 * (index - first_index) + 1));

When to Use This Pattern

When asked for the widest level in a tree including gaps, use BFS with index tracking to measure width efficiently.

Common Mistakes

Mistake: Not normalizing indices by subtracting first_index causing integer overflow
Fix: Normalize indices at each level by subtracting first_index before calculating child indices

Mistake: Counting only existing nodes ignoring gaps, leading to wrong width
Fix: Use indices to count gaps between nodes, not just node count

Summary

Finds the maximum width of a binary tree by counting nodes and gaps at each level.

Use when you need to measure the widest level including empty spaces in a tree.

Track node positions with indices during BFS to count gaps and calculate width accurately.

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