Traversal and Printing a Linked List in DSA Python

Mental Model

We move step-by-step through each connected item in the list to see or print its value.

Analogy: Imagine walking down a line of people holding hands, looking at each person's face one by one.

head -> [1] -> [2] -> [3] -> null
          ↑

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3, print all values

Goal: Print all values in the list from start to end in order

Step 1: Start at the head node with value 1

head -> [1 ↑] -> [2] -> [3] -> null

Why: We begin printing from the first node

Step 2: Print value 1 and move to next node

head -> [1] -> [2 ↑] -> [3] -> null

Why: After printing current node, move forward to continue

Step 3: Print value 2 and move to next node

head -> [1] -> [2] -> [3 ↑] -> null

Why: Keep printing each node until we reach the end

Step 4: Print value 3 and move to next node (null)

head -> [1] -> [2] -> [3] -> null ↑

Why: Last node printed, next is null which ends traversal

Result:

1 -> 2 -> 3 -> null

Annotated Code

DSA Python

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

class LinkedList:
    def __init__(self):
        self.head = None

    def append(self, value):
        new_node = Node(value)
        if not self.head:
            self.head = new_node
            return
        curr = self.head
        while curr.next:
            curr = curr.next
        curr.next = new_node

    def print_list(self):
        curr = self.head
        while curr:
            print(curr.value, end=' -> ')
            curr = curr.next
        print('null')

# Driver code
ll = LinkedList()
ll.append(1)
ll.append(2)
ll.append(3)
ll.print_list()

curr = self.head

start traversal from the first node

while curr:

continue until we reach the end (null)

print(curr.value, end=' -> ')

print current node's value followed by arrow

curr = curr.next

move to next node to continue traversal

Output

1 -> 2 -> 3 -> null

Complexity Analysis

Time: O(n) because we visit each of the n nodes exactly once

Space: O(1) because we only use a few variables regardless of list size

vs Alternative: Compared to recursive printing which uses O(n) space on call stack, this iterative approach is more space efficient

Edge Cases

empty list

prints only 'null' since there are no nodes

DSA Python

while curr:

single node list

prints the single value followed by 'null'

DSA Python

while curr:

When to Use This Pattern

When you need to visit or display every item in a linked list, use traversal by moving from head to next until null.

Common Mistakes

Mistake: Forgetting to move to the next node inside the loop, causing an infinite loop
Fix: Add 'curr = curr.next' inside the while loop to advance traversal

Mistake: Printing 'null' inside the loop instead of after traversal ends
Fix: Print 'null' only after the loop finishes to show list end

Summary

It visits each node in the linked list from start to end and prints its value.

Use it when you want to see or process all elements in order in a linked list.

The key is to start at the head and keep moving to the next node until you reach 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