Right Side View of Binary Tree in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the "Right Side View" of a binary tree?

The right side view of a binary tree is the list of nodes visible when the tree is seen from the right side. It includes the rightmost node at each level of the tree.

beginner

Which traversal method is commonly used to get the right side view of a binary tree?

Level order traversal (Breadth-First Search) is commonly used, where at each level, the last node visited is part of the right side view.

intermediate

In Rust, which data structure is typically used to implement the queue for level order traversal?

A VecDeque is typically used in Rust as a queue for efficient push and pop operations from both ends during level order traversal.

beginner

Why do we pick the last node at each level during level order traversal for the right side view?

Because the last node visited at each level is the rightmost node visible from the right side of the tree.

intermediate

What is the time complexity of finding the right side view of a binary tree?

The time complexity is O(n), where n is the number of nodes, because each node is visited once during the traversal.

What data structure is best suited for level order traversal in Rust?

A VecDeque

B HashMap

C BinaryHeap

D LinkedList

VecDeque allows efficient push and pop from both ends, making it ideal for queue operations in level order traversal.

Which node is included in the right side view at each level?

A The rightmost node

B The leftmost node

C The root node only

D All nodes at that level

The rightmost node at each level is visible from the right side and included in the right side view.

What is the main traversal technique used to find the right side view?

A Depth-First Search

B Level Order Traversal

C Inorder Traversal

D Postorder Traversal

Level order traversal visits nodes level by level, which helps identify the rightmost node at each level.

What is the time complexity of the right side view algorithm?

A O(log n)

B O(n^2)

C O(n)

D O(1)

Each node is visited once, so the time complexity is linear, O(n).

In Rust, which method removes the first element from a VecDeque?

A pop_back()

B remove(0)

C pop()

D pop_front()

pop_front() removes and returns the first element from a VecDeque.

Explain how to find the right side view of a binary tree using level order traversal.

Think about which nodes are visible from the right side at each level.

Describe how you would implement the right side view algorithm in Rust.

Focus on Rust data structures and BFS steps.

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