Right Side View of Binary Tree in DSA Rust - Time & Space Complexity

Time Complexity: Right Side View of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the right side view of a binary tree changes as the tree grows.

How does the number of nodes affect the work done to see the tree from the right side?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn right_side_view(root: Option>) -> Vec {
    let mut result = Vec::new();
    let mut queue = std::collections::VecDeque::new();
    if root.is_some() {
        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 == level_size - 1 {
                result.push(node.val);
            }
            if let Some(left) = node.left {
                queue.push_back(left);
            }
            if let Some(right) = node.right {
                queue.push_back(right);
            }
        }
    }
    result
}
    

This code visits each node level by level and collects the last node's value at each level to form the right side view.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in a breadth-first manner.
• How many times: Each node is processed exactly one time.

How Execution Grows With Input

As the number of nodes grows, the code visits each node once, so the work grows directly with the number of nodes.

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

Pattern observation: The work grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the right side view grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Since we only want the right side nodes, we only visit a few nodes, so it's faster than visiting all."

[OK] Correct: Even to find the rightmost node at each level, we must look at all nodes at that level because the rightmost node depends on the entire level's nodes.

Interview Connect

Understanding this helps you explain how breadth-first search works and how to analyze tree traversal costs, a common skill in interviews.

Self-Check

"What if we changed the traversal to depth-first search instead of breadth-first? How would the time complexity change?"

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