Level-order traversal (BFS) in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the queue with the root node for level-order traversal.

Data Structures Theory

queue = [[1]]

Drag options to blanks, or click blank then click option'

ANone

B[]

Croot

Dstack

Attempts:

3 left

💡 Hint

Common Mistakes

Using an empty list instead of the root node.

Initializing with None instead of the root.

Using a stack instead of a queue.

The level-order traversal starts by adding the root node to the queue.

2fill in blank

medium

Complete the code to remove the first node from the queue during traversal.

Data Structures Theory

current_node = queue.[1]

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Apop

Bpop(0)

Cdequeue

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() which removes the last element.

Using remove() which requires a value, not an index.

Using dequeue() which is not a list method.

In level-order traversal, nodes are removed from the front of the queue using pop(0).

3fill in blank

hard

Fix the error in the code to add the left child to the queue if it exists.

Data Structures Theory

if current_node.[1] is not None:
    queue.append(current_node.left)

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AleftNode

Bchild

Cleft_child

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect attribute names like 'child' or 'left_child'.

Using camelCase instead of snake_case.

The left child of a node is accessed by the attribute 'left'.

4fill in blank

hard

Fill both blanks to add the right child to the queue if it exists.

Data Structures Theory

if current_node.[1] is not None:
    queue.[2](current_node.right)

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Aright

Bappend

Cadd

Dright_child

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'add' instead of 'append' for lists.

Using incorrect attribute names like 'right_child'.

The right child is accessed by 'right' and added to the queue using 'append'.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each node to its level in the tree during BFS.

Data Structures Theory

levels = { [1]: [2] for [3] in nodes }

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Anode.value

Blevel

Cnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.value as key instead of node.

Mixing variable names inconsistently.

This comprehension creates a dictionary where each node maps to its level. 'node' is the variable, 'level' is the value, and 'node.value' would be incorrect here.

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