Push Using Linked List Node in DSA Python

Mental Model

Adding a new item at the start of a linked list means making it the new head and linking it to the old head.

Analogy: Imagine a line of people holding hands. To add a new person at the front, they hold the new person's hand, and the new person holds the old first person's hand.

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

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null, push value 0

Goal: Add the value 0 at the start of the list so it becomes the new head

Step 1: Create new node with value 0

new_node(0) -> null
head -> 1 -> 2 -> 3 -> null

Why: We need a new node to add to the list

Step 2: Set new node's next to current head (node 1)

new_node(0) -> 1 -> 2 -> 3 -> null
head -> 1 -> 2 -> 3 -> null

Why: Link new node to the existing list so it points to the old first node

Step 3: Update head to new node

head -> 0 -> 1 -> 2 -> 3 -> null

Why: Make new node the first node in the list

Result:

head -> 0 -> 1 -> 2 -> 3 -> null

Annotated Code

DSA Python

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

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

    def push(self, new_data):
        new_node = Node(new_data)  # create new node with data
        new_node.next = self.head  # link new node to current head
        self.head = new_node       # update head to new node

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

# Driver code
llist = LinkedList()
llist.push(3)
llist.push(2)
llist.push(1)
llist.push(0)  # push 0 at front
llist.print_list()

new_node = Node(new_data) # create new node with data

create a new node to add at the front

new_node.next = self.head # link new node to current head

point new node's next to current head to keep list connected

self.head = new_node # update head to new node

make new node the new head of the list

Output

0 -> 1 -> 2 -> 3 -> null

Complexity Analysis

Time: O(1) because we only change a few pointers without traversing the list

Space: O(1) because we create only one new node regardless of list size

vs Alternative: Compared to adding at the end which takes O(n) to find the tail, pushing at front is faster and simpler

Edge Cases

empty list

new node becomes the head and points to null

DSA Python

new_node.next = self.head  # link new node to current head

single element list

new node points to the single existing node and becomes new head

DSA Python

self.head = new_node       # update head to new node

When to Use This Pattern

When you need to add an element quickly at the start of a list, use the push pattern to insert at head in constant time.

Common Mistakes

Mistake: Forgetting to update the head pointer to the new node
Fix: Always assign the new node to head after linking it to the old head

Mistake: Not linking the new node's next to the old head, breaking the list
Fix: Set new_node.next = head before updating head

Summary

Adds a new node at the start of a linked list by updating pointers.

Use when you want to insert elements quickly at the front of the list.

The new node points to the old head, then becomes the new head.