Left Side View of Binary Tree in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new binary tree node with given value.

DSA Rust

struct TreeNode {
    val: i32,
    left: Option<Box<TreeNode>>,
    right: Option<Box<TreeNode>>,
};

impl TreeNode {
    fn new(val: i32) -> Self {
        TreeNode {
            val: [1],
            left: None,
            right: None,
        }
    }
};

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

A0

BBox::new(val)

CNone

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of the input value.

Using None which is for options, not for integer values.

The new function should assign the input parameter 'val' to the node's val field.

2fill in blank

medium

Complete the code to add the left child to a TreeNode.

DSA Rust

impl TreeNode {
    fn add_left(&mut self, child: TreeNode) {
        self.left = Some(Box::new([1]));
    }
};

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

Achild

B*child

Cself

DBox::new(child)

Attempts:

3 left

💡 Hint

Common Mistakes

Using Box::new(child) inside Box::new again causes error.

Using *child tries to dereference a non-pointer.

We wrap the given child node in Box::new and then in Some. Here, Box::new is already called outside, so just use 'child'.

3fill in blank

hard

Fix the error in the recursive function to collect left side view nodes.

DSA Rust

fn left_side_view(root: &Option<Box<TreeNode>>, level: usize, result: &mut Vec<i32>) {
    if let Some(node) = root {
        if level == [1] {
            result.push(node.val);
        }
        left_side_view(&node.left, level + 1, result);
        left_side_view(&node.right, level + 1, result);
    }
}

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

Aresult.len()

Blevel

C0

Dnode.val

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing level with 0 always adds only root.

Comparing level with level is always true, causing duplicates.

We add the node's value only if the current level equals the number of nodes already in result, ensuring the first node at each level is recorded.

4fill in blank

hard

Fill both blanks to complete the function that returns the left side view of a binary tree.

DSA Rust

fn left_view(root: Option<Box<TreeNode>>) -> Vec<i32> {
    let mut result = Vec::new();
    [1](&root, 0, &mut result);
    result
}

fn [2](root: &Option<Box<TreeNode>>, level: usize, result: &mut Vec<i32>) {
    if let Some(node) = root {
        if level == result.len() {
            result.push(node.val);
        }
        [2](&node.left, level + 1, result);
        [2](&node.right, level + 1, result);
    }
}

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

Aleft_side_view

Bleft_view

Cleft_side_view_recursive

Dview_left

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names for the helper function causes errors.

Calling the function with wrong parameters.

The helper function is named 'left_side_view' and is called inside 'left_view'.

5fill in blank

hard

Fill all three blanks to create a complete working example that builds a tree and prints its left side view.

DSA Rust

fn main() {
    let mut root = TreeNode::new(1);
    root.add_left(TreeNode::new([1]));
    root.add_right(TreeNode::new([2]));
    let left_view_result = left_view(Some(Box::new(root)));
    println!("Left side view: {:?}", [3]);
}

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

A2

B3

Cleft_view_result

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right child values.

Printing the root instead of the left view result.

We add left child with value 2, right child with value 3, and print the left_view_result vector.

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