Height of Binary Tree in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return 0 when the tree node is empty.

DSA Rust

fn height(node: &Option<Box<TreeNode>>) -> i32 {
    if node.is_none() {
        return [1];
    }
    1
}

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

A0

BNone

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 for empty node

Returning None which is not an integer

When the node is empty (None), the height is 0 because there are no nodes.

2fill in blank

medium

Complete the code to get the left child of the current node.

DSA Rust

fn height(node: &Option<Box<TreeNode>>) -> i32 {
    if node.is_none() {
        return 0;
    }
    let left_height = height(&node.as_ref().unwrap().[1]);
    1
}

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

Aright

Bvalue

Cleft

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using right instead of left

Trying to access value instead of child nodes

To get the left child of the current node, use left field.

3fill in blank

hard

Fix the error in the code to compute the maximum height between left and right subtrees.

DSA Rust

fn height(node: &Option<Box<TreeNode>>) -> i32 {
    if node.is_none() {
        return 0;
    }
    let left_height = height(&node.as_ref().unwrap().left);
    let right_height = height(&node.as_ref().unwrap().right);
    1 + [1](left_height, right_height)
}

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

Amax

Bmin

Csum

Dabs

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max

Adding heights instead of taking max

We take the maximum of left and right subtree heights to find the height of the tree.

4fill in blank

hard

Fill both blanks to complete the recursive height function correctly.

DSA Rust

fn height(node: &Option<Box<TreeNode>>) -> i32 {
    if node.is_none() {
        return [1];
    }
    let left_height = height(&node.as_ref().unwrap().[2]);
    let right_height = height(&node.as_ref().unwrap().right);
    1 + std::cmp::max(left_height, right_height)
}

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

A0

B1

Cleft

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 for empty node

Accessing wrong child field

Return 0 for empty node and access left child with left.

5fill in blank

hard

Fill all three blanks to complete the height function with variable names and return statement.

DSA Rust

fn height([1]: &Option<Box<TreeNode>>) -> i32 {
    if [2].is_none() {
        return 0;
    }
    let left_height = height(&[3].as_ref().unwrap().left);
    let right_height = height(&[3].as_ref().unwrap().right);
    1 + std::cmp::max(left_height, right_height)
}

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

Anode

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently

Using undefined variables

The parameter and variable used is node consistently.

Practice

1. What is the height of a binary tree with only one node (the root)?

easy

A. 1

B. Undefined

C. 2

D. 0

Solution

1. Step 1: Understand height definition

   The height of a binary tree is the number of edges on the longest path from the root to a leaf.

2. Step 2: Apply to single node tree

   With only one node (root), the longest path has length 0 edges.

3. Final Answer:

   0 -> Option D

4. Quick Check:

   Single node tree height = 0 [OK]

Hint: Height counts edges from root to deepest leaf [OK]

Common Mistakes:

• Confusing height with number of nodes
• Thinking empty tree has height 1
• Assuming height starts at 1

2. Which of the following Rust function signatures correctly represents a function to calculate the height of a binary tree node defined as struct Node { left: Option, right: Option }?

easy

A. fn height(node: &Option) -> usize

B. fn height(node: Node) -> usize

C. fn height(node: &Node) -> usize

D. fn height(node: Option<Node>) -> usize

Solution

1. Step 1: Analyze parameter types

   The tree nodes are wrapped in Option to allow nullability and heap allocation.

2. Step 2: Choose correct reference type

   Passing a reference to Option (&Option) avoids moving ownership and allows checking for None.

3. Final Answer:

   fn height(node: &Option) -> usize -> Option A

4. Quick Check:

   Use &Option to handle nullable nodes safely [OK]

Hint: Use &Option to avoid ownership issues [OK]

Common Mistakes:

• Passing Node by value moves ownership
• Using Option<Node> instead of Option
• Using &Node misses None case

3. Given the following Rust code, what is the printed height of the binary tree?

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

fn height(node: &Option<Box<Node>>) -> usize {
    match node {
        None => 0,
        Some(n) => 1 + height(n.left).max(height(n.right)),
    }
}

fn main() {
    let leaf = Node { left: None, right: None };
    let left_child = Node { left: Some(Box::new(leaf)), right: None };
    let root = Some(Box::new(Node { left: Some(Box::new(left_child)), right: None }));
    println!("{}", height(&root));
}

medium

A. 0

B. 2

C. 3

D. 1

Solution

1. Step 1: Trace leaf node height

   The leaf node has no children, so height = 1 (0 + 1).

2. Step 2: Trace left_child height

   Left_child has left = leaf (height 1), right = None (height 0), so height = 1 + max(1,0) = 2.

3. Step 3: Trace root height

   Root has left = left_child (height 2), right = None (height 0), so height = 1 + max(2,0) = 3.

4. Final Answer:

   3 -> Option C

5. Quick Check:

   Height counts nodes on longest path = 3 [OK]

Hint: Count nodes from root to deepest leaf recursively [OK]

Common Mistakes:

• Counting edges instead of nodes
• Ignoring None children
• Misreading recursive calls

4. Identify the bug in this Rust function to calculate binary tree height and select the correct fix:

fn height(node: Option<Box<Node>>) -> usize {
    if node.is_none() {
        return 0;
    }
    let n = node.unwrap();
    1 + height(n.left).max(height(n.right))
}

medium

A. Replace unwrap() with expect() to handle errors

B. Change parameter to &Option<Box<Node>> and use references to avoid moving ownership

C. Return 1 instead of 0 when node is None

D. Use Box<Node> instead of Option<Box<Node>> as parameter

Solution

1. Step 1: Identify ownership issue

   Calling unwrap() moves ownership of node, making recursive calls invalid as ownership is lost.

2. Step 2: Fix by using references

   Changing parameter to &Option<Box<Node>> allows borrowing without moving ownership, enabling safe recursion.

3. Final Answer:

   Change parameter to &Option<Box<Node>> and use references to avoid moving ownership -> Option B

4. Quick Check:

   Use references to prevent ownership errors in recursion [OK]

Hint: Use references to avoid moving nodes in recursion [OK]

Common Mistakes:

• Using unwrap() moves ownership
• Ignoring ownership rules
• Changing return values incorrectly

5. You want to find the height of a binary tree but only have access to a vector of edges representing parent-child relationships as tuples (parent, child). Which approach correctly computes the height of the tree in Rust?

hard

A. Build the tree nodes from edges, then recursively compute height from root node

B. Count the number of edges in the vector and return that as height

C. Use the number of unique nodes as the height

D. Return the maximum child value from edges as height

Solution

1. Step 1: Understand input format

   The vector of edges gives parent-child pairs but no direct tree structure.

2. Step 2: Build tree structure

   Construct nodes and link children to parents to form the tree, identifying the root (node with no parent).

3. Step 3: Compute height recursively

   Starting from root node, recursively calculate height by finding max height of left and right subtrees.

4. Final Answer:

   Build the tree nodes from edges, then recursively compute height from root node -> Option A

5. Quick Check:

   Build tree first, then compute height recursively [OK]

Hint: Convert edges to tree nodes before height calculation [OK]

Common Mistakes:

• Using edge count as height
• Ignoring tree structure
• Assuming max child value equals height