Traversal and Printing a Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move to the next node in the linked list.

DSA Python

current = head
while current is not None:
    print(current.data)
    current = current[1]

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

A.prev

B.head

C.data

D.next

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev instead of .next moves backward, which is incorrect for forward traversal.

Using .data instead of .next tries to assign data to current, causing errors.

The next attribute points to the next node in the linked list. Moving current to current.next traverses the list.

2fill in blank

medium

Complete the code to start traversal from the first node of the linked list.

DSA Python

def print_list(head):
    current = [1]
    while current is not None:
        print(current.data)
        current = current.next

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

Ahead

Bhead.next

CNone

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from head.next skips the first node.

Starting from None means no traversal happens.

Traversal starts from the head node, so current should be set to head.

3fill in blank

hard

Fix the error in the loop condition to correctly traverse the linked list.

DSA Python

current = head
while current[1]None:
    print(current.data)
    current = current.next

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

A!=

B=

Cis

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' assigns instead of compares, causing errors.

Using '==' checks equality but loop should continue while not None.

Using 'is' would check if current is None, preventing traversal.

The loop should continue while current is not None, so the condition must be current != None.

4fill in blank

hard

Fill both blanks to create a dictionary of node data and their positions in the linked list.

DSA Python

positions = {}
index = 0
current = head
while current is not None:
    positions[[1]] = index
    current = [2]
    index += 1

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

Acurrent.data

Bcurrent.next

Cindex

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using index as key stores position as key, which is incorrect here.

Using head instead of current.next causes infinite loop.

Use current.data as the key to store the index. Move to the next node with current.next.

5fill in blank

hard

Fill all three blanks to create a list of node data values from the linked list.

DSA Python

def list_values(head):
    values = []
    current = [1]
    while current is not None:
        values.[2](current.[3])
        current = current.next
    return values

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

Ahead

Bappend

Cdata

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop removes items instead of adding.

Starting from current instead of head causes error if current undefined.

Start from head, add each node's data to values using append.

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