Traversal and Printing a Linked List in DSA Python - Time & Space Complexity

Time Complexity: Traversal and Printing a Linked List

O(n)

Understanding Time Complexity

When we traverse and print a linked list, we want to know how the time taken grows as the list gets longer.

We ask: How does the number of steps change when the list size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

This code goes through each node in the linked list and prints its data.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once per node, so as many times as the number of nodes (n).

How Execution Grows With Input

As the list gets longer, the number of steps grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	10 steps (visiting each node once)
100	100 steps
1000	1000 steps

Pattern observation: The steps increase in a straight line as the list size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to print the list grows directly with the number of nodes.

Common Mistake

[X] Wrong: "Printing a linked list is always constant time because we just start at the head."

[OK] Correct: Even though we start at the head, we must visit every node to print it, so time grows with list size.

Interview Connect

Understanding how traversing a linked list scales helps you explain simple but important operations clearly in interviews.

Self-Check

"What if we only printed every other node? How would the time complexity change?"

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