Left Side View of Binary Tree in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Left Side View for a Simple Binary Tree

What is the output of the left side view for the given binary tree after running the Rust 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 left_side_view(root: Option<Box<TreeNode>>) -> Vec<i32> {
    let mut result = Vec::new();
    if root.is_none() {
        return result;
    }
    let mut queue = VecDeque::new();
    queue.push_back(root);

    while !queue.is_empty() {
        let level_size = queue.len();
        for i in 0..level_size {
            let node_opt = queue.pop_front().unwrap();
            if let Some(node) = node_opt {
                if i == 0 {
                    result.push(node.val);
                }
                if node.left.is_some() {
                    queue.push_back(node.left.clone());
                }
                if node.right.is_some() {
                    queue.push_back(node.right.clone());
                }
            }
        }
    }
    result
}

fn main() {
    let mut root = Box::new(TreeNode::new(1));
    root.left = Some(Box::new(TreeNode::new(2)));
    root.right = Some(Box::new(TreeNode::new(3)));
    root.left.as_mut().unwrap().left = Some(Box::new(TreeNode::new(4)));
    root.right.as_mut().unwrap().right = Some(Box::new(TreeNode::new(5)));

    let view = left_side_view(Some(root));
    println!("{:?}", view);
}

A [1, 2, 5]

B [1, 2, 4]

C [1, 3, 5]

D [1, 3, 4]

💡 Hint

The left side view shows the first node visible at each level from the left side.

The left side view visits nodes level by level and picks the first node at each level. Here, level 0 has 1, level 1 has 2 (left child), and level 2 has 4 (left child of 2).

🧠 Conceptual

intermediate

Understanding the Left Side View Concept

Which statement best describes the left side view of a binary tree?

A It shows the nodes visible when the tree is viewed from the left side, picking the first node at each level.

B It shows all leaf nodes of the binary tree.

C It shows the nodes visible when the tree is viewed from the right side, picking the last node at each level.

D It shows the nodes in the tree in in-order traversal.

💡 Hint

Think about what you see if you stand on the left side of the tree.

The left side view is the set of nodes visible when looking at the tree from the left side, which means the first node encountered at each level.

🔧 Debug

advanced

Identify the Bug in Left Side View Implementation

What error will occur when running the following Rust code for left side view of a binary tree?

DSA Rust

fn left_side_view(root: Option<Box<TreeNode>>) -> Vec<i32> {
    let mut result = Vec::new();
    let mut queue = VecDeque::new();
    queue.push_back(root.unwrap());

    while !queue.is_empty() {
        let level_size = queue.len();
        for i in 0..level_size {
            let node = queue.pop_front().unwrap();
            if i == 0 {
                result.push(node.val);
            }
            if node.left.is_some() {
                queue.push_back(node.left.unwrap());
            }
            if node.right.is_some() {
                queue.push_back(node.right.unwrap());
            }
        }
    }
    result
}

A Panic at unwrap on empty Option if root is None

B Incorrect output because it pushes right child before left child

C Infinite loop because queue is never emptied

D Compilation error due to missing use of VecDeque

💡 Hint

Check how unwrap is used on Option values.

If root is None, calling unwrap() on it causes a panic at runtime. The code does not check if root is None before unwrapping.

❓ Predict Output

advanced

Output of Left Side View for Unbalanced Binary Tree

What is the output of the left side view function for this unbalanced binary tree?

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 left_side_view(root: Option<Box<TreeNode>>) -> Vec<i32> {
    let mut result = Vec::new();
    if root.is_none() {
        return result;
    }
    let mut queue = VecDeque::new();
    queue.push_back(root);

    while !queue.is_empty() {
        let level_size = queue.len();
        for i in 0..level_size {
            let node_opt = queue.pop_front().unwrap();
            if let Some(node) = node_opt {
                if i == 0 {
                    result.push(node.val);
                }
                if node.left.is_some() {
                    queue.push_back(node.left.clone());
                }
                if node.right.is_some() {
                    queue.push_back(node.right.clone());
                }
            }
        }
    }
    result
}

fn main() {
    let mut root = Box::new(TreeNode::new(10));
    root.right = Some(Box::new(TreeNode::new(20)));
    root.right.as_mut().unwrap().right = Some(Box::new(TreeNode::new(30)));
    root.right.as_mut().unwrap().right.as_mut().unwrap().left = Some(Box::new(TreeNode::new(40)));

    let view = left_side_view(Some(root));
    println!("{:?}", view);
}

A [10, 30, 40]

B [10, 20]

C [10, 20, 30, 40]

D [10, 20, 40]

💡 Hint

Remember the left side view picks the first node at each level from left to right.

At level 0: 10; level 1: 20 (only right child); level 2: 30 (right child of 20); level 3: 40 (left child of 30). All are first nodes at their levels.

🧠 Conceptual

expert

Time Complexity of Left Side View Algorithm

What is the time complexity of the standard breadth-first search approach to find the left side view of a binary tree with n nodes?

A O(1), because it only stores the first node of each level

B O(log n), because it only visits nodes on the left side

C O(n^2), because it processes each level separately

D O(n), because each node is visited exactly once

💡 Hint

Consider how many nodes the algorithm visits in total.

The algorithm visits every node once in a breadth-first manner, so the time complexity is linear in the number of nodes.

Practice

1. What does the Left Side View of a binary tree represent?

easy

A. All leaf nodes of the tree

B. Nodes visible when the tree is viewed from the right side

C. Nodes visible when the tree is viewed from the left side

D. Nodes at the deepest level of the tree

Solution

1. Step 1: Understand the view perspective

   The left side view shows nodes visible when looking at the tree from the left side.

2. Step 2: Identify visible nodes

   Only the first node at each level from the left is visible in this view.

3. Final Answer:

   Nodes visible when the tree is viewed from the left side -> Option C

4. Quick Check:

   Left side view = nodes seen from left [OK]

Hint: Think of standing on the left side and looking at the tree [OK]

Common Mistakes:

• Confusing left side view with right side view
• Including all leaf nodes instead of first nodes per level
• Thinking it shows deepest nodes only

2. Which Rust code snippet correctly declares a binary tree node struct for left side view traversal?

easy

A. struct TreeNode { val: i32, left: TreeNode, right: TreeNode }

B. struct TreeNode { val: i32, left: Box<TreeNode>, right: Box<TreeNode> }

C. struct TreeNode { val: i32, left: &TreeNode, right: &TreeNode }

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

Solution

1. Step 1: Check ownership and optional children

   Rust binary trees use Option for nullable child nodes.

2. Step 2: Validate correct syntax

   Option allows for null or Some(child), which is correct for left and right.

3. Final Answer:

   struct TreeNode { val: i32, left: Option, right: Option } -> Option D

4. Quick Check:

   Use Option for nullable children [OK]

Hint: Use Option for child nodes in Rust [OK]

Common Mistakes:

• Using direct TreeNode type for children (causes infinite size)
• Using references without lifetimes
• Omitting Option for nullable children

3. What is the output of the left side view function for this tree?

    1
   / \
  2   3
   \   \
    5   4

Assuming the function returns a vector of node values visible from the left side.

medium

A. [1, 3, 4]

B. [1, 2, 5]

C. [1, 2, 4]

D. [1, 5, 4]

Solution

1. Step 1: Identify nodes visible from left at each level

   Level 0: 1, Level 1: 2 (left child), Level 2: 5 (right child of 2)

2. Step 2: Collect visible nodes in order

   Left side view nodes are [1, 2, 5]

3. Final Answer:

   [1, 2, 5] -> Option B

4. Quick Check:

   Left side nodes = [1, 2, 5] [OK]

Hint: Pick first node at each level from left [OK]

Common Mistakes:

• Choosing right child nodes instead of left
• Including nodes not visible from left
• Mixing order of nodes

4. Given this Rust function snippet for left side view, what is the bug?

fn left_side_view(root: Option) -> Vec<i32> {
    let mut result = vec![];
    let mut queue = vec![root.unwrap()];
    while !queue.is_empty() {
        let level_size = queue.len();
        for i in 0..level_size {
            let node = queue.remove(0);
            if i == 0 {
                result.push(node.val);
            }
            if let Some(left) = node.left {
                queue.push(left);
            }
            if let Some(right) = node.right {
                queue.push(right);
            }
        }
    }
    result
}

medium

A. Calling unwrap() on root without checking if it is null

B. Using remove(0) on vector causes runtime error

C. Pushing right child before left child breaks left view order

D. Not returning the result vector at the end

Solution

1. Step 1: Check root unwrap usage

   Calling unwrap() on root without checking if it is null can cause panic if tree is empty.

2. Step 2: Confirm other parts are correct

   remove(0) works but is inefficient; order of pushing left then right is correct; result is returned.

3. Final Answer:

   Calling unwrap() on root without checking if it is null -> Option A

4. Quick Check:

   unwrap() on null causes panic [OK]

Hint: Check unwrap() calls for null before use [OK]

Common Mistakes:

• Ignoring unwrap panic possibility
• Thinking remove(0) causes error (just slow)
• Confusing child push order
• Missing return statement

5. You want to modify the left side view function to return the leftmost node values only for levels where the node value is even. Which approach is correct?

hard

A. Add a check inside the level loop: if i == 0 and node.val % 2 == 0, then push node.val to result

B. Push node.val for all nodes during traversal, then filter even values from result after traversal

C. Push node.val only if node.val is even, ignoring position in level

D. Push node.val if it is even and it is the rightmost node at the level

Solution

1. Step 1: Understand requirement for leftmost even nodes

   We want only the first node at each level if its value is even.

2. Step 2: Apply condition during traversal

   Check if i == 0 (leftmost) and node.val % 2 == 0 before pushing to result.

3. Final Answer:

   Add a check inside the level loop: if i == 0 and node.val % 2 == 0, then push node.val to result -> Option A

4. Quick Check:

   Filter leftmost nodes by even value during traversal [OK]

Hint: Check leftmost and even condition together inside loop [OK]

Common Mistakes:

• Filtering after traversal loses position info
• Ignoring leftmost position and filtering only by value
• Checking rightmost node instead of leftmost