Traversal and Printing a Linked List in DSA Python - Build from Scratch

Traversal and Printing a Linked List

📖 Scenario: Imagine you have a chain of boxes, each holding a number and a link to the next box. This is like a linked list. You want to see all the numbers in order by opening each box one by one.

🎯 Goal: You will build a simple linked list with three nodes, then write code to go through the list from start to end and print all the values in order.

📋 What You'll Learn

Create a linked list with three nodes containing values 10, 20, and 30

Use a variable called head to point to the first node

Traverse the linked list using a variable called current

Print the values of all nodes in order separated by ' -> ' and ending with ' -> None'

💡 Why This Matters

🌍 Real World

Linked lists are used in many software systems to manage ordered data where items can be added or removed easily, like music playlists or undo history.

💼 Career

Understanding linked lists is fundamental for software developers, especially when working with low-level data structures or optimizing memory usage.

1

Create the linked list nodes

Create a class called Node with an __init__ method that takes value and sets self.value and self.next to None. Then create three nodes called node1, node2, and node3 with values 10, 20, and 30 respectively.

DSA Python

# Define the Node class and create three nodes
# Your code here

Hint

Start by defining the Node class with an __init__ method. Then create three nodes with the given values.

2

Link the nodes together

Set node1.next to node2 and node2.next to node3. Then create a variable called head and set it to node1.

DSA Python

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

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)

# Link the nodes and set head
# Your code here

Hint

Connect the first node to the second, the second to the third, and set head to the first node.

3

Traverse the linked list

Create a variable called current and set it to head. Use a while loop that runs while current is not None. Inside the loop, print current.value followed by ' -> ' without a newline, then update current to current.next.

DSA Python

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

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)

node1.next = node2
node2.next = node3
head = node1

# Traverse the list and print values
# Your code here

Hint

Use a while loop to go through each node until you reach the end (None). Print each node's value followed by ' -> '.

4

Print the end of the list

After the while loop, print 'None' to show the end of the linked list.

DSA Python

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

node1 = Node(10)
node2 = Node(20)
node3 = Node(30)

node1.next = node2
node2.next = node3
head = node1

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

# Print the end of the list
# Your code here

Hint

After printing all node values, print 'None' to show the list ends here.

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