Stack Implementation Using Linked List in DSA C# - Execution Trace

Concept Flow - Stack Implementation Using Linked List

Create new node with value

↓

Set new node.next to current top

↓

Update top pointer to new node

↓

Stack push complete

↓

Check if top is null?

Yes↓

Stack is empty

No↓

Read value from top node

↓

Update top to top.next

↓

Stack pop complete

Push adds a new node at the top by linking it to the current top. Pop removes the top node and updates the top pointer.

Execution Sample

DSA C#

Push(10)
Push(20)
Pop()
Push(30)
Pop()

This code pushes 10, then 20, pops the top (20), pushes 30, then pops the top (30).

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Push(10)	10	top -> new node(10)	10 -> null
2	Push(20)	20 -> 10	new node(20).next = top(10), top -> new node(20)	20 -> 10 -> null
3	Pop()	10	top -> top.next (10)	10 -> null
4	Push(30)	30 -> 10	new node(30).next = top(10), top -> new node(30)	30 -> 10 -> null
5	Pop()	10	top -> top.next (10)	10 -> null
6	End	10	No change	10 -> null

💡 No more operations; stack ends with one node containing 10.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
top	null	node(10)	node(20)	node(10)	node(30)	node(10)	node(10)
stack size	0	1	2	1	2	1	1

Key Moments - 3 Insights

Why does push set new node's next to current top before updating top?

Because the new node must point to the previous top to keep the stack linked. See step 2 in execution_table where new node(20).next points to node(10).

What happens to the popped node after pop operation?

The popped node is no longer referenced by top, so it becomes unreachable and can be cleaned up by memory management. See step 3 where top moves from node(20) to node(10).

Why does pop update top to top.next instead of removing the node directly?

Because top.next points to the next node in the stack. Updating top to top.next effectively removes the top node from the stack. See step 3 and 5 in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what is the visual state of the stack?

A 10 -> 20 -> null

B 20 -> 10 -> null

C 20 -> null

D 10 -> null

💡 Hint

Check the 'Visual State' column at step 2 in execution_table.

At which step does the top pointer move from node(20) to node(10)?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'Pointer Changes' column to see when top updates to top.next.

If we add Push(40) after step 5, what will be the new visual state?

A 40 -> 10 -> null

B 10 -> 40 -> null

C 30 -> 40 -> 10 -> null

D 40 -> 30 -> 10 -> null

💡 Hint

Push adds new node at top pointing to current top; see steps 2 and 4 for pattern.

Concept Snapshot

Stack using linked list:
- Push: create node, set node.next = top, top = node
- Pop: top = top.next
- Top points to stack's top node
- Stack grows/shrinks dynamically
- No fixed size limit

Full Transcript

This visualization shows how a stack is built using a linked list. Each push creates a new node and links it to the current top, then updates top to this new node. Each pop removes the top node by moving top to the next node. The execution table tracks the nodes in the list, pointer changes, and the visual state after each operation. Variable tracker shows how the top pointer and stack size change. Key moments clarify why the new node points to the old top and how pop removes nodes. The quiz tests understanding of stack state and pointer updates.

Practice

1. What is the main advantage of implementing a stack using a linked list instead of an array?

easy

A. It uses less memory than an array

B. The stack can grow dynamically without a fixed size limit

C. It allows random access to elements

D. It automatically sorts the elements

Solution

1. Step 1: Understand stack size limitation in arrays

   Arrays have a fixed size, so stacks implemented with arrays have a maximum capacity.

2. Step 2: Linked list dynamic size

   Linked lists can grow or shrink dynamically by adding or removing nodes, so the stack size is limited only by memory.

3. Final Answer:

   The stack can grow dynamically without a fixed size limit -> Option B

4. Quick Check:

   Linked list stacks grow dynamically [OK]

Hint: Linked list stacks grow dynamically, arrays do not [OK]

Common Mistakes:

• Thinking linked list uses less memory always
• Assuming linked list allows random access
• Confusing stack sorting with implementation

2. Which of the following is the correct way to push an element onto a stack implemented using a linked list in C#?

easy

A. Create a new node, set its next to current top, then update top to new node

B. Create a new node, set its previous to current top, then update top to new node

C. Add the new element at the end of the linked list

D. Replace the top node's value with the new element

Solution

1. Step 1: Understand push operation in linked list stack

   Push adds a new node at the beginning (top) of the linked list.

2. Step 2: Correct pointer update

   The new node's next points to the current top, then top is updated to new node.

3. Final Answer:

   Create a new node, set its next to current top, then update top to new node -> Option A

4. Quick Check:

   Push adds node at front [OK]

Hint: Push adds node at front, update top pointer [OK]

Common Mistakes:

• Using previous pointer instead of next
• Adding node at end instead of front
• Overwriting top node value instead of adding new node

3. What will be the output after the following C# code executes?

StackNode top = null;
// Push 10
StackNode node1 = new StackNode(10);
node1.next = top;
top = node1;
// Push 20
StackNode node2 = new StackNode(20);
node2.next = top;
top = node2;
// Pop
int popped = top.data;
top = top.next;
// Print top data
Console.WriteLine(top.data);

medium

A. NullReferenceException

B. 20

C. 10

D. 0

Solution

1. Step 1: Trace push operations

   Push 10: top points to node with data 10.
   Push 20: top points to node with data 20, next points to node 10.

2. Step 2: Trace pop operation

   Pop removes top node (20), top updated to next node (10).

3. Step 3: Print top data

   Top now points to node with data 10, so output is 10.

4. Final Answer:

   10 -> Option C

5. Quick Check:

   After pop, top data = 10 [OK]

Hint: Pop removes top, next node becomes new top [OK]

Common Mistakes:

• Printing popped value instead of new top
• Confusing push order
• Not updating top after pop

4. Identify the error in this pop method for a stack implemented using a linked list:

public int Pop() {
    if (top == null) {
        Console.WriteLine("Stack Underflow");
    }
    int value = top.data;
    top = top.next;
    return value;
}

medium

A. It does not return a value when stack is empty

B. It does not update the top pointer

C. It incorrectly prints "Stack Overflow" instead of "Underflow"

D. It removes the bottom node instead of the top

Solution

1. Step 1: Check behavior when stack is empty

   If top is null, method prints message but continues to access top.data causing error or returns no value.

2. Step 2: Missing return for empty stack case

   Method should return or throw exception when stack is empty to avoid runtime error.

3. Final Answer:

   It does not return a value when stack is empty -> Option A

4. Quick Check:

   Empty stack must return or throw [OK]

Hint: Always return or throw on empty stack pop [OK]

Common Mistakes:

• Ignoring empty stack case
• Confusing overflow with underflow
• Not updating top pointer

5. You want to implement a method ReverseStack that reverses the elements of a stack implemented using a linked list without using extra data structures. Which approach is correct?

hard

A. Use a queue to dequeue and enqueue elements to reverse order

B. Traverse the linked list and swap data values of nodes pairwise

C. Create a new linked list and push elements from old stack to new stack

D. Use recursion to pop all elements, then push them back in reverse order

Solution

1. Step 1: Understand constraints

   No extra data structures allowed, so using recursion (call stack) is allowed but no explicit new lists or queues.

2. Step 2: Recursion approach

   Recursively pop all elements, then insert them back at bottom to reverse order without extra storage.

3. Step 3: Other options use extra structures

   Swapping data or using new lists or queues violates constraints or is incorrect for stack reversal.

4. Final Answer:

   Use recursion to pop all elements, then push them back in reverse order -> Option D

5. Quick Check:

   Recursion reverses stack without extra structures [OK]

Hint: Use recursion to reverse stack without extra data structures [OK]

Common Mistakes:

• Using extra linked lists or queues
• Trying to swap node data incorrectly
• Ignoring no extra structure constraint