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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a vector to store the right side view nodes.

DSA Rust

let mut right_view = Vec::[1]();

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Adefault

Bnew

Cwith_capacity

Dfrom

Attempts:

3 left

💡 Hint

Common Mistakes

Using Vec::with_capacity without specifying capacity.

Using Vec::from which requires an initial collection.

We use Vec::new() to create an empty vector to store the right side view nodes.

2fill in blank

medium

Complete the code to push the rightmost node's value into the result vector.

DSA Rust

right_view.[1](node.val);

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Apush

Binsert

Cappend

Dextend

Attempts:

3 left

💡 Hint

Common Mistakes

Using insert without specifying index.

Using append or extend which expect collections.

We use push to add a single element to the end of the vector.

3fill in blank

hard

Fix the error in the recursive call to traverse the right subtree first.

DSA Rust

if let Some(right) = node.[1] {
    dfs(&right, depth + 1, right_view);
}

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Aleft

Bnode

Cchild

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using left child instead of right child.

Using an invalid field name.

To get the right side view, we must traverse the right child first, so we access node.right.

4fill in blank

hard

Fill both blanks to check if current depth equals the vector length and then push the node's value.

DSA Rust

if depth == right_view.[1]() {
    right_view.[2](node.val);
}

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Alen

Bpush

Ccapacity

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using capacity instead of len.

Using insert without index.

We compare depth with right_view.len() to see if this is the first node at this depth, then push the value.

5fill in blank

hard

Fill all three blanks to complete the DFS function signature and call.

DSA Rust

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

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

Adepth

Bright_view

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing variable names for depth and vector.

Passing immutable references instead of mutable.

The DFS function takes depth as the current level and right_view as the mutable vector to store values. The call uses these parameters accordingly.

Practice

1. What does the "Right Side View of Binary Tree" represent?

easy

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

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

C. All leaf nodes of the binary tree

D. All nodes at the deepest level of the tree

Solution

1. Step 1: Understand the view perspective

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

2. Step 2: Differentiate from other views

   Unlike left side or leaf views, it only includes the rightmost nodes at each level.

3. Final Answer:

   Nodes visible when the tree is viewed from the right side -> Option A

4. Quick Check:

   Right side view = rightmost nodes [OK]

Hint: Right side view shows rightmost nodes at each level [OK]

Common Mistakes:

• Confusing right side view with left side view
• Including all leaf nodes instead of rightmost nodes
• Thinking it shows all nodes at the deepest level

2. Which Rust code snippet correctly declares a binary tree node struct for the right side view problem?

easy

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

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

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

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

Solution

1. Step 1: Check ownership and optional children

   Rust needs Option for nullable child nodes with ownership.

2. Step 2: Validate correct struct syntax

   Option allows recursive tree structure safely.

3. Final Answer:

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

4. Quick Check:

   Use Option for children [OK]

Hint: Use Option for nullable child nodes [OK]

Common Mistakes:

• Using direct TreeNode type causing infinite size error
• Using references without lifetimes
• Using Option<TreeNode> without Box causing compile error

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

    1
   / \
  2   3
   \   \
    5   4

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

medium

A. [1, 2, 4]

B. [1, 2, 5]

C. [1, 3, 4]

D. [1, 3, 5]

Solution

1. Step 1: Identify rightmost nodes at each level

   Level 0: 1, Level 1: 3 (right child), Level 2: 4 (right child of 3)

2. Step 2: Collect these nodes in order

   Right side view vector is [1, 3, 4]

3. Final Answer:

   [1, 3, 4] -> Option C

4. Quick Check:

   Rightmost nodes per level = [1, 3, 4] [OK]

Hint: Pick rightmost node at each tree level [OK]

Common Mistakes:

• Choosing left child nodes instead of rightmost
• Including nodes not visible from right side
• Mixing order of nodes in output

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

fn right_side_view(root: Option) -> Vec<i32> {
    let mut result = Vec::new();
    let mut queue = Vec::new();
    if root.is_some() {
        queue.push(root.unwrap());
    }
    while !queue.is_empty() {
        let level_size = queue.len();
        for i in 0..level_size {
            let node = queue.remove(0);
            if i == level_size - 1 {
                result.push(node.val);
            }
            if node.left.is_some() {
                queue.push(node.left.unwrap());
            }
            if node.right.is_some() {
                queue.push(node.right.unwrap());
            }
        }
    }
    result
}

medium

A. The function returns the left side view, not right side

B. Queue should be a linked list, not a vector

C. The loop should iterate backwards

D. Using unwrap() consumes the Option, causing ownership errors

Solution

1. Step 1: Analyze use of unwrap() on Option

   Calling unwrap() moves the value out, so subsequent uses cause ownership errors.

2. Step 2: Identify correct approach

   Use references or cloning to avoid moving ownership when pushing children into queue.

3. Final Answer:

   Using unwrap() consumes the Option, causing ownership errors -> Option D

4. Quick Check:

   unwrap() moves ownership, causing errors [OK]

Hint: Avoid unwrap() that moves ownership inside loops [OK]

Common Mistakes:

• Ignoring ownership rules in Rust
• Thinking vector is wrong data structure here
• Misunderstanding loop direction relevance

5. How would you modify a standard level order traversal in Rust to return the right side view of a binary tree efficiently?

hard

A. Traverse each level left to right, record last node's value per level

B. Traverse each level right to left, record first node's value per level

C. Traverse each level left to right, record first node's value per level

D. Traverse each level right to left, record last node's value per level

Solution

1. Step 1: Understand level order traversal direction

   Standard level order goes left to right per level.

2. Step 2: Identify which node to record

   Recording the last node at each level gives the rightmost visible node.

3. Final Answer:

   Traverse each level left to right, record last node's value per level -> Option A

4. Quick Check:

   Last node per level in left-to-right traversal = right side view [OK]

Hint: Record last node at each level in left-to-right traversal [OK]

Common Mistakes:

• Recording first node instead of last
• Changing traversal direction unnecessarily
• Confusing left and right side views