Level-order traversal (BFS) in Data Structures Theory - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding the order of nodes visited in BFS

Consider a binary tree where the root node is 1, its left child is 2, and right child is 3. Node 2 has children 4 and 5, and node 3 has children 6 and 7.

What is the order of nodes visited in a level-order traversal (BFS)?

A 1, 2, 3, 4, 5, 6, 7

B 1, 3, 2, 7, 6, 5, 4

C 4, 5, 6, 7, 2, 3, 1

D 1, 2, 4, 5, 3, 6, 7

💡 Hint

Level-order traversal visits nodes level by level from left to right.

In BFS, nodes are visited starting from the root, then all nodes at the next level from left to right, and so on. So the order is 1 (root), then 2 and 3 (level 2), then 4, 5, 6, 7 (level 3).

📋 Factual

intermediate

Queue usage in BFS

Which data structure is primarily used to implement level-order traversal (BFS) on a tree?

A Stack

B Hash Map

C Queue

D Priority Queue

💡 Hint

Think about the order nodes are visited and how to keep track of nodes to visit next.

BFS uses a queue to keep track of nodes to visit in the order they are discovered, ensuring nodes are processed level by level.

🚀 Application

advanced

Number of nodes visited at each level

Given a binary tree with 15 nodes perfectly balanced (each node has 0 or 2 children), how many nodes will BFS visit at level 3 (root is level 1)?

A 8

B 4

C 2

D 1

💡 Hint

Count nodes level by level in a perfect binary tree.

In a perfect binary tree, level 1 has 1 node, level 2 has 2 nodes, level 3 has 4 nodes, and level 4 has 8 nodes. Since the tree has 15 nodes, it has 4 levels. So level 3 has 4 nodes.

🔍 Analysis

advanced

Time complexity of BFS on a tree

What is the time complexity of performing a level-order traversal (BFS) on a tree with n nodes?

A O(n log n)

B O(log n)

C O(n^2)

D O(n)

💡 Hint

Consider how many times each node is visited and processed.

BFS visits each node exactly once and processes its children, so the total work is proportional to the number of nodes, which is O(n).

❓ Reasoning

expert

Effect of adding cycles on BFS traversal

Suppose you apply a BFS algorithm designed for trees on a graph that contains cycles but do not mark visited nodes. What will happen?

A The BFS will enter an infinite loop visiting nodes repeatedly.

B The BFS will visit each node exactly once as in a tree.

C The BFS will skip some nodes due to cycles.

D The BFS will terminate immediately without visiting any nodes.

💡 Hint

Think about what happens when nodes are revisited in a graph with cycles.

Without marking visited nodes, BFS will keep revisiting nodes in cycles endlessly, causing an infinite loop.

Practice

1. What is the main data structure used in level-order traversal (BFS) of a binary tree?

easy

A. Linked List

B. Stack

C. Queue

D. Hash Table

Solution

1. Step 1: Understand traversal method

   Level-order traversal visits nodes level by level, which requires processing nodes in the order they appear.

2. Step 2: Identify suitable data structure

   A queue follows First-In-First-Out (FIFO) order, perfect for visiting nodes level by level.

3. Final Answer:

   Queue -> Option C

4. Quick Check:

   Level-order traversal uses a queue [OK]

Hint: Level-order uses FIFO structure: queue [OK]

Common Mistakes:

• Confusing queue with stack (LIFO)
• Thinking hash table stores order
• Assuming linked list is used directly

2. Which of the following is the correct syntax to enqueue a node n into a queue named q in Python during level-order traversal?

easy

A. q.append(n)

B. q.enqueue(n)

C. q.push(n)

D. q.insert(n)

Solution

1. Step 1: Identify Python queue implementation

   In Python, a list can be used as a queue where append() adds elements to the end.

2. Step 2: Confirm enqueue operation

   Using q.append(n) correctly adds node n to the queue's rear.

3. Final Answer:

   q.append(n) -> Option A

4. Quick Check:

   Python queue enqueue uses append() [OK]

Hint: Use append() to add nodes to Python queue [OK]

Common Mistakes:

• Using push() which is not a Python list method
• Using enqueue() which is not built-in
• Using insert() which adds at wrong position

3. Given the binary tree:

    1
   / \
  2   3
 /   / \
4   5   6

What is the output of a level-order traversal?

medium

A. [1, 3, 2, 6, 5, 4]

B. [1, 2, 3, 4, 5, 6]

C. [4, 2, 5, 3, 6, 1]

D. [1, 2, 4, 3, 5, 6]

Solution

1. Step 1: Traverse level by level

   Start at root: 1. Then next level: 2 and 3. Then next level: 4, 5, 6.

2. Step 2: List nodes in visiting order

   Collect nodes as visited: [1, 2, 3, 4, 5, 6].

3. Final Answer:

   [1, 2, 3, 4, 5, 6] -> Option B

4. Quick Check:

   Level-order visits nodes top to bottom, left to right [OK]

Hint: Visit nodes level by level, left to right [OK]

Common Mistakes:

• Mixing order of nodes in same level
• Listing nodes in depth-first order
• Reversing levels incorrectly

4. Consider this Python snippet for level-order traversal:

queue = [root]
while queue:
    node = queue.pop()
    print(node.value)
    if node.left:
        queue.append(node.left)
    if node.right:
        queue.append(node.right)

What is the main error in this code?

medium

A. Using pop() removes last element, not first

B. Appending children before popping node

C. Not checking if node is null

D. Printing node value before adding children

Solution

1. Step 1: Understand queue behavior

   Level-order traversal requires FIFO order, so nodes must be removed from the front.

2. Step 2: Identify pop() behavior

   pop() without index removes last element (LIFO), causing incorrect traversal order.

3. Final Answer:

   Using pop() removes last element, not first -> Option A

4. Quick Check:

   pop() removes from end, use pop(0) for queue [OK]

Hint: Use pop(0) to dequeue from front in Python list [OK]

Common Mistakes:

• Using pop() instead of pop(0)
• Ignoring queue order importance
• Assuming append order fixes pop issue

5. You want to find the shortest path from the root to a target node in a binary tree using level-order traversal. Which modification ensures you stop traversal as soon as the target is found?

hard

A. Continue traversal until queue is empty, then check target

B. Traverse only left children until target is found

C. Add all nodes to a stack and pop until target is found

D. Check each node during dequeue; stop and return path if target found

Solution

1. Step 1: Understand BFS for shortest path

   BFS visits nodes level by level, so the first time target is found is the shortest path.

2. Step 2: Implement early stopping

   Check each node when dequeued; if it matches target, stop traversal immediately and return path.

3. Final Answer:

   Check each node during dequeue; stop and return path if target found -> Option D

4. Quick Check:

   Stop BFS on target found for shortest path [OK]

Hint: Stop BFS immediately when target node is dequeued [OK]

Common Mistakes:

• Traversing entire tree unnecessarily
• Using stack instead of queue for shortest path
• Ignoring right children in traversal