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

Recall & Review

beginner

What is the Left Side View of a Binary Tree?

It is the list of nodes visible when the tree is viewed from the left side. For each level, only the first node encountered from the left is included.

beginner

Which tree traversal method is commonly used to find the Left Side View of a Binary Tree?

Level-order traversal (Breadth-First Search) is commonly used because it processes nodes level by level, allowing us to pick the first node at each level.

intermediate

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

A VecDeque is used because it allows efficient insertion and removal from both ends, which is ideal for queue operations.

beginner

Why do we only record the first node at each level for the Left Side View?

Because the first node at each level is the leftmost node visible from the left side. Other nodes at the same level are hidden behind it.

intermediate

What is the time complexity of finding the Left 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 HashMap

B BinaryHeap

C VecDeque

D LinkedList

VecDeque allows efficient queue operations needed for level-order traversal.

Which node is included in the Left Side View at each level?

A The rightmost node

B The leftmost node

C The deepest node

D The root node only

The leftmost node at each level is visible from the left side.

What traversal method is used to get the Left Side View?

A Level-order traversal

B Pre-order traversal

C In-order traversal

D Post-order traversal

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

What is the time complexity of the Left Side View algorithm?

A O(n)

B O(log n)

C O(n^2)

D O(1)

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

If a binary tree has 5 levels, how many nodes will the Left Side View contain at most?

A 1

B Depends on the number of nodes

C 10

D 5

At most one node per level is included, so 5 levels mean up to 5 nodes.

Explain how to find the Left Side View of a Binary Tree using level-order traversal.

Think about how you see the tree from the left side and how BFS helps.

Describe the role of VecDeque in implementing the Left Side View algorithm in Rust.

Consider why a queue is needed for BFS and why VecDeque is preferred in Rust.

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