Traversal and Printing a Linked List in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main purpose of traversing a linked list?

To visit each node in the linked list one by one, usually to read or print the data stored in each node.

beginner

In a singly linked list, how do you move from one node to the next?

By following the 'next' pointer/reference stored in the current node to reach the next node.

beginner

What condition tells you that you have reached the end of a linked list during traversal?

When the 'next' pointer/reference of the current node is None (or null in other languages), it means the end of the list.

beginner

Why is it important to start traversal from the head node of a linked list?

Because the head node is the entry point to the list; without it, you cannot access the other nodes.

beginner

What is the output format when printing a linked list like 1 -> 2 -> 3 -> null?

It shows the sequence of node values connected by arrows, ending with 'null' to indicate the list ends.

What does the 'next' pointer in a linked list node represent?

A The size of the list

B The data stored in the node

C The previous node in the list

D The address of the next node in the list

The 'next' pointer holds the address/reference to the next node, allowing traversal.

How do you know when to stop traversing a linked list?

A When the head node is reached

B After visiting 10 nodes

C When the current node's next pointer is null

D When the current node's data is zero

Traversal stops when the next pointer is null, meaning no more nodes.

Which node do you start with when printing a linked list?

A The head node

B The last node

C Any random node

D The middle node

Traversal and printing always start from the head node.

What is the output of printing a linked list with nodes containing 5, 10, 15?

A null -> 5 -> 10 -> 15

B 5 -> 10 -> 15 -> null

C 15 -> 10 -> 5 -> null

D 5, 10, 15

The list is printed from head to tail with arrows and ends with null.

What happens if you try to traverse a linked list starting from a null head?

A Nothing is printed because the list is empty

B It prints all nodes

C It causes an error

D It prints null repeatedly

A null head means the list is empty, so no nodes to print.

Explain step-by-step how to traverse and print all elements of a singly linked list.

Think about visiting each node one by one until the end.

Describe how the linked list structure supports traversal and printing of its elements.

Focus on the role of the next pointer and head node.

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