Creating a Singly Linked List from Scratch in DSA Python - Step-by-Step

Mental Model

A singly linked list is a chain of boxes where each box holds a value and points to the next box in line.

Analogy: Imagine a treasure hunt where each clue leads you to the next clue. Each clue is like a box holding a value and a pointer to the next clue.

head -> [value | next] -> [value | next] -> [value | null]
↑

Dry Run Walkthrough

Input: Create a singly linked list with values 1, 2, 3 in order

Goal: Build a linked list where each node points to the next, ending with null

Step 1: Create first node with value 1 and set it as head

head -> [1 | null]
↑

Why: We need a starting point for the list

Step 2: Create second node with value 2 and link it after first node

head -> [1 | next] -> [2 | null]

Why: Link second node so list continues after first

Step 3: Create third node with value 3 and link it after second node

head -> [1 | next] -> [2 | next] -> [3 | null]

Why: Link third node to complete the chain

Result:

head -> 1 -> 2 -> 3 -> null

Annotated Code

DSA Python

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

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

    def append(self, value):
        new_node = Node(value)  # create new node
        if not self.head:
            self.head = new_node  # set head if list empty
            return
        curr = self.head
        while curr.next:
            curr = curr.next  # move to last node
        curr.next = new_node  # link new node at end

    def __str__(self):
        values = []
        curr = self.head
        while curr:
            values.append(str(curr.value))
            curr = curr.next  # move to next node
        return ' -> '.join(values) + ' -> null'

# Driver code
linked_list = SinglyLinkedList()
linked_list.append(1)
linked_list.append(2)
linked_list.append(3)
print(linked_list)

new_node = Node(value) # create new node

create a new box with the given value

if not self.head: self.head = new_node # set head if list empty return

if list is empty, start it with new node

while curr.next: curr = curr.next # move to last node

walk through nodes until last one

curr.next = new_node # link new node at end

attach new node after last node

while curr: values.append(str(curr.value)) curr = curr.next # move to next node

collect all node values to print list

Output

1 -> 2 -> 3 -> null

Complexity Analysis

Time: O(n) because to add a node at the end, we may need to walk through all n nodes

Space: O(n) because we store n nodes in memory

vs Alternative: Compared to arrays, linked lists allow easy insertion without shifting elements but require walking nodes to find the end

Edge Cases

Empty list when appending first node

The new node becomes the head of the list

DSA Python

if not self.head:
    self.head = new_node  # set head if list empty

When to Use This Pattern

When you need a flexible list that grows by linking nodes instead of fixed-size arrays, use singly linked list creation.

Common Mistakes

Mistake: Not setting the head when the list is empty
Fix: Check if head is None and assign new node as head before traversing

Summary

Creates a chain of nodes where each node points to the next, forming a singly linked list.

Use when you want a list structure that can grow dynamically without resizing arrays.

The key is to keep track of the head and link each new node at the end by updating pointers.

Practice

1. What is the main purpose of a Singly Linked List in data structures?

easy

A. To store elements with keys and values for fast lookup

B. To store elements in a sequence where each element points to the next

C. To store elements in a sorted array for binary search

D. To store elements in a tree structure with parent and child nodes

Solution

1. Step 1: Understand the structure of a singly linked list

   A singly linked list stores elements in nodes where each node points to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike arrays or trees, singly linked lists do not store keys or have parent-child relationships; they only link sequentially.

3. Final Answer:

   To store elements in a sequence where each element points to the next -> Option B

4. Quick Check:

   Singly Linked List = sequence with next pointers [OK]

Hint: Remember: singly linked list nodes point only forward [OK]

Common Mistakes:

• Confusing linked lists with arrays or hash maps
• Thinking nodes have backward pointers (that's doubly linked list)
• Assuming linked lists store key-value pairs

2. Which of the following is the correct way to define a node class for a singly linked list in Python?

easy

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

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

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

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

Solution

1. Step 1: Identify the correct attributes for a singly linked list node

   A singly linked list node stores data and a pointer to the next node only, so it needs data and next.

2. Step 2: Check the constructor parameters and attributes

   class Node: def __init__(self, data): self.data = data self.next = None correctly initializes data and sets next to None by default. Other options include prev which is for doubly linked lists or incorrect parameters.

3. Final Answer:

   class Node: def __init__(self, data): self.data = data self.next = None -> Option C

4. Quick Check:

   Node has data and next only [OK]

Hint: Singly linked list nodes have no 'prev' pointer [OK]

Common Mistakes:

• Adding a 'prev' pointer by mistake
• Not initializing 'next' to None
• Using extra parameters unnecessarily

3. What will be the output of the following code that creates a singly linked list with nodes 10, 20, 30 and prints the list?

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

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

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

    def print_list(self):
        current = self.head
        while current:
            print(current.data, end=' -> ')
            current = current.next
        print('None')

ll = LinkedList()
ll.append(10)
ll.append(20)
ll.append(30)
ll.print_list()

medium

A. 10 -> 20 -> None

B. 30 -> 20 -> 10 -> None

C. None

D. 10 -> 20 -> 30 -> None

Solution

1. Step 1: Trace the append method calls

   Appending 10 sets head to node with data 10. Then 20 and 30 are added at the end by traversing the list.

2. Step 2: Trace the print_list method

   Prints nodes in order: 10, then 20, then 30, each followed by ' -> ', ending with 'None'.

3. Final Answer:

   10 -> 20 -> 30 -> None -> Option D

4. Quick Check:

   Append adds at end, print shows order [OK]

Hint: Append adds nodes at list end; print shows order [OK]

Common Mistakes:

• Assuming append adds at front
• Printing without 'None' at end
• Confusing order of nodes

4. The following code tries to append nodes to a singly linked list but does not work as expected. What is the error?

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

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

    def append(self, data):
        new_node = Node(data)
        if self.head is None:
            self.head = new_node
        else:
            last = self.head
            while last.next:
                last = last.next
            last = new_node

ll = LinkedList()
ll.append(1)
ll.append(2)
ll.append(3)

current = ll.head
while current:
    print(current.data, end=' -> ')
    current = current.next
print('None')

medium

A. The line 'last = new_node' does not link the new node to the list

B. The head is never initialized

C. The Node class is missing the 'prev' pointer

D. The while loop condition should be 'while last is not None'

Solution

1. Step 1: Analyze the append method's else block

   The code finds the last node correctly but then does 'last = new_node' which only changes the local variable, not the last node's next pointer.

2. Step 2: Identify the fix

   To link the new node, it should be 'last.next = new_node' to update the last node's next pointer.

3. Final Answer:

   The line 'last = new_node' does not link the new node to the list -> Option A

4. Quick Check:

   Assign to last.next to link nodes [OK]

Hint: Update last.next, not last variable, to link nodes [OK]

Common Mistakes:

• Assigning new_node to last instead of last.next
• Not updating the head when list is empty
• Confusing variable assignment with pointer update

5. You want to create a singly linked list from scratch that supports inserting a new node at the beginning and printing the list. Which code snippet correctly implements this?

hard

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

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

def insert_at_beginning(self, data):
new_node = Node(data)
new_node.next = self.head
self.head = new_node

def print_list(self):
current = self.head
while current:
print(current.data, end=' -> ')
current = current.next
print('None')

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

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

def insert_at_beginning(self, data):
new_node = Node(data)
new_node.prev = self.head
self.head = new_node

def print_list(self):
current = self.head
while current:
print(current.data, end=' -> ')
current = current.next
print('None')

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

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

def insert_at_beginning(self, data):
new_node = Node(data)
self.head = new_node

def print_list(self):
current = self.head
while current:
print(current.data, end=' -> ')
current = current.next
print('None')

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

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

def insert_at_beginning(self, data):
new_node = Node(data)
new_node.next = None
self.head = new_node

def print_list(self):
current = self.head
while current:
print(current.data, end=' -> ')
current = current.next
print('None')

Solution

1. Step 1: Understand insert at beginning logic

   To insert at the beginning, new node's next should point to current head, then head updated to new node.

2. Step 2: Check each option's insert_at_beginning method

   class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def insert_at_beginning(self, data): new_node = Node(data) new_node.next = self.head self.head = new_node def print_list(self): current = self.head while current: print(current.data, end=' -> ') current = current.next print('None') correctly sets new_node.next = self.head and updates head. class Node: def __init__(self, data): self.data = data self.prev = None self.next = None class LinkedList: def __init__(self): self.head = None def insert_at_beginning(self, data): new_node = Node(data) new_node.prev = self.head self.head = new_node def print_list(self): current = self.head while current: print(current.data, end=' -> ') current = current.next print('None') wrongly uses prev pointer. class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def insert_at_beginning(self, data): new_node = Node(data) self.head = new_node def print_list(self): current = self.head while current: print(current.data, end=' -> ') current = current.next print('None') misses linking new_node.next. class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None def insert_at_beginning(self, data): new_node = Node(data) new_node.next = None self.head = new_node def print_list(self): current = self.head while current: print(current.data, end=' -> ') current = current.next print('None') sets new_node.next to None, losing old list.

3. Final Answer:

   Option A code snippet correctly inserts at beginning and prints list -> Option A

4. Quick Check:

   New node points to old head, then head updated [OK]

Hint: New node's next = old head; update head to new node [OK]

Common Mistakes:

• Not linking new node to old head
• Using prev pointer in singly linked list
• Setting new_node.next to None incorrectly