Why Push Using Linked List Node in DSA Python?

The Big Idea

What if adding something on top could be done in just one quick step, no matter how big your stack is?

The Scenario

Imagine you have a stack of books on your desk. You want to add a new book on top, but you have to move all the books one by one to make space. This takes a lot of time and effort.

The Problem

Manually shifting all items to add one at the top is slow and easy to mess up. You might drop a book or lose track of the order. It's tiring and error-prone.

The Solution

Using a linked list node to push means you just create a new node and link it to the current top. No shifting needed. It's quick, clean, and keeps the order safe.

Before vs After

✗ Before

stack = [1, 2, 3]
new_item = 4
stack = [new_item] + stack  # manually create new list

✓ After

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

new_node = Node(4)
new_node.next = top
 top = new_node

What It Enables

This lets you add items instantly on top, making stacks fast and reliable for many tasks.

Real Life Example

When you undo typing in a text editor, each action is pushed onto a stack using linked list nodes, so you can quickly reverse your steps.

Key Takeaways

Manual shifting is slow and risky.

Linked list nodes let you add on top instantly.

This keeps data safe and operations fast.

Practice

1. What does the push operation do in a linked list when using a node?

easy

A. Removes the last node from the linked list

B. Searches for a node with a specific value

C. Deletes the first node of the linked list

D. Adds a new node at the beginning of the linked list

Solution

1. Step 1: Understand push operation meaning

   Push means adding a new element to the start of the linked list.

2. Step 2: Identify push effect on linked list

   The new node becomes the head, pointing to the old head node.

3. Final Answer:

   Adds a new node at the beginning of the linked list -> Option D

4. Quick Check:

   Push = Add at start [OK]

Hint: Push always adds node at the start of the list [OK]

Common Mistakes:

• Thinking push adds at the end
• Confusing push with pop (removal)
• Assuming push deletes nodes

2. Which of the following is the correct syntax to create a new node and push it to the front of a linked list in Python?

easy

A. head.next = Node(data); head = head.next

B. head = Node(data); head.next = new_node

C. new_node = Node(data); new_node.next = head; head = new_node

D. new_node = Node(data); head = new_node.next

Solution

1. Step 1: Create new node with data

   Use new_node = Node(data) to create a node.

2. Step 2: Link new node to current head and update head

   Set new_node.next = head and then head = new_node.

3. Final Answer:

   new_node = Node(data); new_node.next = head; head = new_node -> Option C

4. Quick Check:

   Correct push syntax = new_node = Node(data); new_node.next = head; head = new_node [OK]

Hint: New node points to head, then head updates to new node [OK]

Common Mistakes:

• Assigning head before linking new_node.next
• Confusing order of assignments
• Using undefined variables

3. What will be the linked list after executing the following code?

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

def push(head, data):
    new_node = Node(data)
    new_node.next = head
    head = new_node
    return head

head = None
head = push(head, 10)
head = push(head, 20)
head = push(head, 30)

medium

A. 30 -> 20 -> 10 -> null

B. 10 -> 20 -> 30 -> null

C. 30 -> 10 -> 20 -> null

D. 20 -> 30 -> 10 -> null

Solution

1. Step 1: Push 10 to empty list

   List becomes 10 -> null.

2. Step 2: Push 20 to front

   New head is 20, pointing to 10: 20 -> 10 -> null.

3. Step 3: Push 30 to front

   New head is 30, pointing to 20: 30 -> 20 -> 10 -> null.

4. Final Answer:

   30 -> 20 -> 10 -> null -> Option A

5. Quick Check:

   Push adds at front, last pushed is first [OK]

Hint: Last pushed node is always the head [OK]

Common Mistakes:

• Assuming push adds at end
• Not updating head after push
• Mixing order of nodes

4. Find the error in this push function for a linked list:

def push(head, data):
    new_node = Node(data)
    head = new_node
    new_node.next = head
    return head

medium

A. new_node is not created properly

B. new_node.next is set after head is updated, causing a loop

C. head is not returned

D. Function does not accept head as parameter

Solution

1. Step 1: Analyze assignment order

   head is set to new_node before linking new_node.next.

2. Step 2: Identify pointer mistake

   new_node.next = head sets next to itself, creating a loop.

3. Final Answer:

   new_node.next is set after head is updated, causing a loop -> Option B

4. Quick Check:

   Link new_node.next before updating head [OK]

Hint: Set new_node.next before changing head [OK]

Common Mistakes:

• Assigning head before linking next
• Creating self-referential loops
• Forgetting to return head

5. Given a linked list with nodes 5 -> 15 -> 25 -> null, what will be the list after pushing nodes with values 35 and 45 in order using the push function?

head = Node(5)
head.next = Node(15)
head.next.next = Node(25)
head = push(head, 35)
head = push(head, 45)

hard

A. 45 -> 35 -> 5 -> 15 -> 25 -> null

B. 5 -> 15 -> 25 -> 35 -> 45 -> null

C. 35 -> 45 -> 5 -> 15 -> 25 -> null

D. 45 -> 35 -> 25 -> 15 -> 5 -> null

Solution

1. Step 1: Original list is 5 -> 15 -> 25 -> null

   Head points to 5, which points to 15, then 25.

2. Step 2: Push 35

   New node 35 points to head (5), head updates to 35: 35 -> 5 -> 15 -> 25 -> null.

3. Step 3: Push 45

   New node 45 points to head (35), head updates to 45: 45 -> 35 -> 5 -> 15 -> 25 -> null.

4. Final Answer:

   45 -> 35 -> 5 -> 15 -> 25 -> null -> Option A

5. Quick Check:

   Push adds nodes at front in order [OK]

Hint: New nodes become new head, pushing old nodes back [OK]

Common Mistakes:

• Appending instead of pushing at front
• Mixing node order after pushes
• Not updating head after push