Traversal and Printing a Linked List in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Traversing a Linked List

What is the printed output of the linked list after running this traversal code?

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

head = Node(1)
head.next = Node(2)
head.next.next = Node(3)

current = head
while current:
    print(current.data, end=' -> ')
    current = current.next
print('null')

A 1 -> 2 -> 3 -> null

B null

C 3 -> 2 -> 1 -> null

D 1 -> 2 -> null

💡 Hint

Think about how the traversal moves from the head node to the end.

The code starts at the head node and prints each node's data followed by '->'. It stops when current is None, then prints 'null'. So the output is '1 -> 2 -> 3 -> null'.

❓ Predict Output

intermediate

What happens if the list is empty?

What is the output when the linked list is empty (head is None) and the same traversal code runs?

DSA Python

head = None
current = head
while current:
    print(current.data, end=' -> ')
    current = current.next
print('null')

A null

B 0 -> null

C None -> null

D Error: AttributeError

💡 Hint

If head is None, does the loop run?

Since head is None, current is None, so the while loop does not run. Only 'null' is printed.

🔧 Debug

advanced

Identify the error in this linked list traversal code

What error does this code produce when run?

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

head = Node(1)
head.next = Node(2)

current = head
while current.next:
    print(current.data, end=' -> ')
    current = current.next
print('null')

A Prints '1 -> 2 -> null'

B Prints '1 -> null'

C Prints '1 -> 2' without 'null'

D Prints '1 -> null' and misses last node data

💡 Hint

Check the loop condition and what nodes get printed.

The loop runs while current.next is not None, so it stops before the last node. The last node's data (2) is not printed, only '1 -> null' is printed.

🧠 Conceptual

advanced

How many nodes are printed by this traversal?

Given a linked list with 5 nodes, what is the number of nodes printed by this code?

DSA Python

current = head
count = 0
while current:
    print(current.data, end=' -> ')
    current = current.next
    count += 1
print('null')
print(count)

A 6

B 5

C 4

D 0

💡 Hint

The loop runs once per node until current is None.

The loop visits each of the 5 nodes exactly once, so count is 5.

🚀 Application

expert

What is the output after reversing and printing the linked list?

What is the printed output after reversing this linked list and then printing it?

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def reverse(head):
    prev = None
    current = head
    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node
    return prev

head = Node(1)
head.next = Node(2)
head.next.next = Node(3)

head = reverse(head)
current = head
while current:
    print(current.data, end=' -> ')
    current = current.next
print('null')

A null

B 1 -> 2 -> 3 -> null

C 3 -> 2 -> 1 -> null

D Error: infinite loop

💡 Hint

Reversing changes the direction of links between nodes.

The reverse function flips the list order. Printing after reversal shows nodes in reverse order: 3, 2, 1, then null.

Practice

1. What does it mean to traverse a linked list?

easy

A. To reverse the order of nodes in the list

B. To add a new node at the end of the list

C. To visit each node in the list one by one from start to end

D. To delete the first node of the list

Solution

1. Step 1: Understand traversal meaning

   Traversal means visiting each element in order to read or process it.

2. Step 2: Apply to linked list context

   In a linked list, traversal means starting at the head and moving through each node until the end.

3. Final Answer:

   To visit each node in the list one by one from start to end -> Option C

4. Quick Check:

   Traversal = Visit nodes sequentially [OK]

Hint: Traversal means visiting nodes one by one [OK]

Common Mistakes:

• Confusing traversal with insertion or deletion
• Thinking traversal changes the list order
• Assuming traversal skips nodes

2. Which of the following is the correct syntax to start traversing a linked list in Python if the head node is named head?

easy

A. current = head
while current != head:
print(current.data)
current = current.next

B. current = head
while current.next is not None:
print(current.data)
current = current.next

C. current = head
for i in range(len(head)):
print(current.data)
current = current.next

D. current = head
while current is not None:
print(current.data)
current = current.next

Solution

1. Step 1: Understand traversal loop condition

   We must continue while current node exists (not None) to visit all nodes.

2. Step 2: Check each option's loop condition

   current = head while current is not None: print(current.data) current = current.next uses while current is not None, which correctly visits all nodes including the last one.

3. Final Answer:

   current = head while current is not None: print(current.data) current = current.next -> Option D

4. Quick Check:

   Loop until current is None [OK]

Hint: Traverse while current is not None [OK]

Common Mistakes:

• Stopping before last node by checking current.next
• Using for loop with range on linked list
• Loop condition that never ends or skips nodes

3. What will be the output of the following code?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

head = Node(10)
head.next = Node(20)
head.next.next = Node(30)

current = head
while current is not None:
    print(current.data, end=' -> ')
    current = current.next
print('None')

medium

A. 10 -> 20 -> 30 -> None

B. 10 20 30 None

C. None -> 10 -> 20 -> 30

D. 10 -> 20 -> None

Solution

1. Step 1: Trace the linked list nodes

   Nodes contain 10, 20, 30 linked in order: 10 -> 20 -> 30 -> None.

2. Step 2: Follow the print loop

   Loop prints each node's data followed by ' -> ', then prints 'None' after loop ends.

3. Final Answer:

   10 -> 20 -> 30 -> None -> Option A

4. Quick Check:

   Print nodes with arrows ending with None [OK]

Hint: Print each node data with arrow until None [OK]

Common Mistakes:

• Missing 'None' at the end
• Printing nodes without arrows
• Stopping before last node

4. The following code is intended to print all elements of a linked list, but it causes an error. What is the error?

current = head
while current.next is not None:
    print(current.data)
    current = current.next
print(current.data)

medium

A. It causes an infinite loop

B. It misses printing the last node inside the loop

C. It raises an AttributeError

D. It prints nodes in reverse order

Solution

1. Step 1: Analyze loop condition

   The loop runs while current.next is not None, so it stops before the last node.

2. Step 2: Check printing after loop

   After loop, it prints the last node's data separately, so all nodes are printed but last node outside loop.

3. Final Answer:

   It misses printing the last node inside the loop -> Option B

4. Quick Check:

   Loop condition skips last node printing [OK]

Hint: Use while current is not None to include last node [OK]

Common Mistakes:

• Using current.next in loop condition
• Expecting all nodes printed inside loop
• Confusing infinite loop with this error

5. You want to print a linked list in reverse order without modifying the list or using extra space like a stack. Which approach below correctly achieves this using recursion?

def print_reverse(node):
    if node is None:
        return
    # What goes here?
    print(node.data, end=' -> ')

hard

A. print_reverse(node.next)

B. print(node.data)
print_reverse(node.next)

C. print_reverse(node)
print(node.data)

D. print_reverse(node.next)
print_reverse(node)

Solution

1. Step 1: Understand recursion for reverse print

   Recursively call function on next node first to reach end before printing.

2. Step 2: Place recursive call before print

   Calling print_reverse(node.next) before printing current node prints in reverse order.

3. Final Answer:

   print_reverse(node.next) -> Option A

4. Quick Check:

   Recursive call before print reverses order [OK]

Hint: Call recursion on next node before printing current [OK]

Common Mistakes:

• Printing before recursive call (prints forward)
• Calling recursion on same node (infinite recursion)
• Calling recursion twice causing errors