Tree Traversal Level Order BFS in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is Level Order Traversal in a tree?

Level Order Traversal visits nodes level by level from top to bottom and left to right within each level, using a queue to keep track of nodes.

beginner

Which data structure is commonly used to implement Level Order Traversal (BFS) in trees?

A queue is used to hold nodes of the current level before visiting their children, ensuring nodes are processed in the correct order.

intermediate

In Rust, which standard library type is best suited for implementing a queue for BFS?

The VecDeque type from std::collections is ideal because it allows efficient push and pop operations from both ends.

beginner

What is the time complexity of Level Order Traversal on a binary tree with n nodes?

The time complexity is O(n) because each node is visited exactly once.

intermediate

Why does Level Order Traversal use BFS instead of DFS?

Because BFS visits nodes level by level, it naturally fits Level Order Traversal, while DFS goes deep into one branch before visiting others.

What data structure is essential for Level Order Traversal?

A Stack

B HashMap

C Queue

D Linked List

Level Order Traversal uses a queue to process nodes in FIFO order.

In Level Order Traversal, which node is visited first?

A Leftmost leaf node

B Root node

C Rightmost leaf node

D Any random node

Traversal starts from the root node and proceeds level by level.

What is the output order of nodes in Level Order Traversal?

A Top to bottom, left to right

B Bottom to top, right to left

C Left subtree first, then right subtree

D Right subtree first, then left subtree

Nodes are visited level by level from top to bottom and left to right.

Which Rust collection is best for BFS queue?

A VecDeque

B HashSet

C BinaryHeap

D LinkedList

VecDeque supports efficient push/pop from both ends, ideal for BFS.

What is the time complexity of Level Order Traversal?

A O(1)

B O(log n)

C O(n^2)

D O(n)

Each node is visited once, so complexity is linear in number of nodes.

Explain how Level Order Traversal (BFS) works on a binary tree.

Think about visiting nodes in layers from top to bottom.

Describe how you would implement Level Order Traversal in Rust.

Focus on queue operations and visiting nodes.

Practice

1. What does the Level Order Traversal (BFS) of a binary tree do?

easy

A. Visits nodes in preorder: root, left, right

B. Visits nodes level by level from top to bottom, left to right

C. Visits nodes in postorder: left, right, root

D. Visits nodes in inorder: left, root, right

Solution

1. Step 1: Understand Level Order Traversal

   Level Order Traversal visits nodes one level at a time, starting from the root and moving downwards.

2. Step 2: Compare with other traversals

   Preorder, inorder, and postorder are depth-first traversals, not level by level.

3. Final Answer:

   Visits nodes level by level from top to bottom, left to right -> Option B

4. Quick Check:

   Level Order = BFS = top to bottom, left to right [OK]

Hint: Level order means visiting nodes level by level [OK]

Common Mistakes:

• Confusing BFS with DFS traversals
• Thinking preorder is level order
• Mixing traversal orders

2. Which Rust code snippet correctly initializes a queue for BFS traversal of a binary tree?

easy

A. let mut queue: VecDeque<&TreeNode> = VecDeque::new();

B. let mut queue: Vec<&TreeNode> = Vec::new();

C. let mut queue: LinkedList<TreeNode> = LinkedList::new();

D. let mut queue: HashMap = HashMap::new();

Solution

1. Step 1: Identify the correct queue type for BFS

   BFS requires a queue with efficient push and pop from front, VecDeque is ideal in Rust.

2. Step 2: Check other options

   Vec is a stack-like structure, LinkedList is less common and less efficient, HashMap is a key-value store, not a queue.

3. Final Answer:

   let mut queue: VecDeque<&TreeNode> = VecDeque::new(); -> Option A

4. Quick Check:

   VecDeque is the queue for BFS in Rust [OK]

Hint: Use VecDeque for queue operations in Rust BFS [OK]

Common Mistakes:

• Using Vec which is inefficient for queue
• Using HashMap instead of queue
• Confusing LinkedList with VecDeque

3. What is the output of the following Rust code performing level order traversal?

use std::collections::VecDeque;

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

fn level_order(root: Option) -> Vec<i32> {
    let mut result = Vec::new();
    let mut queue = VecDeque::new();
    if let Some(node) = root {
        queue.push_back(node);
    }
    while let Some(node) = queue.pop_front() {
        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
}

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode { val: 2, left: None, right: None })),
        right: Some(Box::new(TreeNode { val: 3, left: None, right: None })),
    }));
    let output = level_order(root);
    println!("{:?}", output);
}

medium

A. [1, 2, 3]

B. [1, 3, 2]

C. [2, 1, 3]

D. [3, 2, 1]

Solution

1. Step 1: Trace the queue operations

   Start with root 1 in queue. Pop 1, add 2 and 3 to queue. Result: [1]

2. Step 2: Continue traversal

   Pop 2, no children. Result: [1, 2]. Pop 3, no children. Result: [1, 2, 3]. Queue empty, stop.

3. Final Answer:

   [1, 2, 3] -> Option A

4. Quick Check:

   Level order visits 1 then 2 then 3 [OK]

Hint: Follow queue push/pop order carefully [OK]

Common Mistakes:

• Mixing left and right child order
• Confusing preorder with level order
• Forgetting to push children into queue

4. Identify the bug in this Rust BFS traversal code snippet:

while let Some(node) = queue.pop_front() {
    result.push(node.val);
    queue.push_back(node.left.unwrap());
    queue.push_back(node.right.unwrap());
}

medium

A. Using pop_front() instead of pop_back()

B. Not pushing root node into queue before loop

C. Result vector is not initialized

D. Using unwrap() directly causes panic if Option is None

Solution

1. Step 1: Analyze unwrap() usage

   Calling unwrap() on Option without safe check can panic if None.

2. Step 2: Check safer alternative

   Better to use if let Some(left) = node.left to safely extract without panic.

3. Final Answer:

   Using unwrap() directly causes panic if Option is None -> Option D

4. Quick Check:

   unwrap() on None panics, use if let Some() [OK]

Hint: Avoid unwrap() on Option without safe check [OK]

Common Mistakes:

• Assuming unwrap() is always safe
• Ignoring initialization of queue
• Confusing pop_front and pop_back

5. Given a binary tree, how would you modify the level order BFS traversal in Rust to return a vector of vectors, where each inner vector contains the node values at that level?

hard

A. Traverse the tree recursively and collect nodes in a single vector, then split by level afterwards

B. Use a stack instead of queue to reverse the order of nodes per level

C. Use a queue and track the number of nodes at each level, collecting values into inner vectors before moving to next level

D. Store all nodes in a HashMap with level as key, then flatten values into a single vector

Solution

1. Step 1: Understand level grouping in BFS

   Track how many nodes are in the current level by queue length before processing that level.

2. Step 2: Collect nodes per level

   For each level, pop exactly that many nodes, collect their values into a vector, then push their children for next level.

3. Final Answer:

   Use a queue and track the number of nodes at each level, collecting values into inner vectors before moving to next level -> Option C

4. Quick Check:

   Level order + count nodes per level = vector of vectors [OK]

Hint: Count nodes per level in queue to group values [OK]

Common Mistakes:

• Using recursion instead of BFS for level grouping
• Using stack which reverses order
• Trying to split flat vector after traversal