Height of Binary Tree in DSA Rust - Time & Space Complexity

Time Complexity: Height of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how long it takes to find the height of a binary tree as the tree grows.

How does the time needed change when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn height(node: &Option>) -> i32 {
    match node {
        None => 0,
        Some(n) => {
            let left_height = height(&n.left);
            let right_height = height(&n.right);
            1 + left_height.max(right_height)
        }
    }
}
    

This code finds the height of a binary tree by checking the height of left and right subtrees recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

As the number of nodes grows, the function visits each node once to calculate height.

Input Size (n)	Approx. Operations
10	About 10 calls
100	About 100 calls
1000	About 1000 calls

Pattern observation: The number of operations grows directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the height grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The height function only checks one path, so it runs in O(log n)."

[OK] Correct: The function visits every node to find the max height, so it must check all paths, not just one.

Interview Connect

Understanding this helps you explain how recursive tree algorithms work and why visiting all nodes matters.

Self-Check

"What if the tree was a linked list (all nodes have only one child)? How would the time complexity change?"

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