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

Concept Flow - Stack Implementation Using Linked List

Create new node

↓

Push operation: new_node.next = top

↓

Update top = new_node

↓

Stack grows with new top

↓

Pop operation: if top != null

↓

Save top node

↓

Update top = top.next

↓

Delete saved node

↓

Stack shrinks with new top

The stack grows by adding new nodes at the top linked to the previous top. Pop removes the top node and updates the top pointer.

Execution Sample

DSA C++

struct Node {
  int data;
  Node* next;
};

// Push 10, Push 20, Pop

This code pushes 10, then 20 onto the stack, then pops the top element.

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), delete node(20)	10 -> null
4	End	10	No change	10 -> null

💡 No more operations, stack top points to node with 10

Variable Tracker

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

Key Moments - 3 Insights

Why does push update new_node.next to top before changing top?

Because in step 2 of execution_table, new_node(20).next is set to current top(10) to link the new node before updating top to new_node(20). This keeps the stack order correct.

What happens to the popped node after pop?

In step 3, the popped node(20) is deleted after top moves to next node(10), freeing memory and removing it from the stack.

Why does the visual state show '-> null' at the end?

Because the last node's next pointer is null, indicating the bottom of the stack with no further nodes.

Visual Quiz - 3 Questions

Look at the execution_table at step 2, what is the top node's data?

A 10

B 20

C null

D 30

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns at step 2

At which step does the stack size decrease?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'stack size' row in variable_tracker after each step

If we push 30 after step 3, what will be the new visual state?

A 30 -> 10 -> null

B 10 -> 30 -> null

C 20 -> 30 -> 10 -> null

D 30 -> null

💡 Hint

Recall push adds new node at top pointing to previous top

Concept Snapshot

Stack using linked list:
- Push: create node, new_node.next = top, top = new_node
- Pop: if top != null, top = top.next, delete old top
- Top points to newest element
- Stack grows/shrinks at top only
- Bottom node next is null

Full Transcript

This visualization shows how a stack is built using a linked list. Each push creates a new node that points to the current top, then updates top to this new node. Each pop removes the top node by moving top to the next node and deleting the old top. The stack grows and shrinks only at the top, maintaining last-in-first-out order. The visual state shows the linked nodes with arrows ending in null at the bottom. Variable tracking shows how top pointer and stack size change after each operation. Key moments clarify why pointers update in a specific order and what happens to popped nodes. The quiz tests understanding of top node data, stack size changes, and push effects.

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 is easier to sort the stack elements

Solution

1. Step 1: Understand stack size limitation in arrays

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

2. Step 2: Linked list dynamic size property

   Linked lists can grow or shrink dynamically by adding or removing nodes, so stack size is not fixed.

3. Final Answer:

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

4. Quick Check:

   Linked list stack size = dynamic growth [OK]

Hint: Linked list stacks grow dynamically, arrays have fixed size [OK]

Common Mistakes:

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

2. Which of the following is the correct syntax to create a new node in a linked list stack in C++?

easy

A. Node* newNode = new Node(value);

B. Node newNode = Node(value);

C. Node newNode = new Node(value);

D. Node* newNode = Node(value);

Solution

1. Step 1: Understand dynamic memory allocation in C++

   To create a node dynamically, use 'new' keyword and pointer type.

2. Step 2: Check syntax correctness

   Node* newNode = new Node(value); uses pointer and 'new' correctly: Node* newNode = new Node(value);

3. Final Answer:

   Node* newNode = new Node(value); -> Option A

4. Quick Check:

   Dynamic node creation = new + pointer [OK]

Hint: Use 'new' with pointer to create nodes dynamically [OK]

Common Mistakes:

• Omitting pointer (*) when using new
• Using stack allocation instead of dynamic
• Assigning new object to non-pointer variable

3. What will be the output after executing the following code snippet for a stack implemented using linked list?

Stack s;
s.push(10);
s.push(20);
s.push(30);
s.pop();
s.push(40);
s.display();

Assuming display prints elements from top to bottom.

medium

A. 40 -> 30 -> 20 -> 10 -> null

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

C. 20 -> 10 -> null

D. 40 -> 20 -> 10 -> null

Solution

1. Step 1: Trace push and pop operations

   Push 10, stack: 10; push 20, stack: 20->10; push 30, stack: 30->20->10; pop removes 30, stack: 20->10; push 40, stack: 40->20->10.

2. Step 2: Determine display output

   Display prints from top to bottom: 40 -> 20 -> 10 -> null.

3. Final Answer:

   40 -> 20 -> 10 -> null -> Option D

4. Quick Check:

   Stack after ops = 40->20->10 [OK]

Hint: Track push/pop stepwise to find top element [OK]

Common Mistakes:

• Forgetting pop removes top element
• Assuming push adds at bottom
• Mixing order of elements in display

4. Identify the error in the following pop function of a linked list stack:

int pop() {
  if (top == NULL) {
    cout << "Stack Underflow";
    return -1;
  }
  int val = top->data;
  delete top;
  top = top->next;
  return val;
}

medium

A. Returning -1 on empty stack is incorrect

B. Missing check for stack overflow

C. Accessing top->next after deleting top causes undefined behavior

D. Not updating top pointer after pop

Solution

1. Step 1: Analyze order of operations in pop

   Code deletes top node before accessing top->next, which is unsafe.

2. Step 2: Identify correct order

   Must save top->next in a temp pointer before deleting top, then update top.

3. Final Answer:

   Accessing top->next after deleting top causes undefined behavior -> Option C

4. Quick Check:

   Delete before next access = error [OK]

Hint: Save next pointer before deleting current node [OK]

Common Mistakes:

• Deleting node before updating pointer
• Ignoring stack underflow check
• Returning wrong values on empty stack

5. You want to implement a function to reverse a stack implemented using a linked list without using extra data structures. Which approach is correct?

hard

A. Use recursion to pop all elements and push them back in reverse order

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

C. Create a new linked list and copy elements in reverse order

D. Use a queue to reorder elements and then push back to stack

Solution

1. Step 1: Understand constraints

   No extra data structures allowed, so cannot use queue or new list.

2. Step 2: Use recursion to reverse stack

   Recursion can pop all elements, then push them back in reverse order naturally.

3. Step 3: Why other options fail

   Swapping data pairwise does not reverse entire stack; copying needs extra list; queue is extra structure.

4. Final Answer:

   Use recursion to pop all elements and push them back in reverse order -> Option A

5. Quick Check:

   Recursion reverses stack without extra structures [OK]

Hint: Recursion can reverse stack without extra memory [OK]

Common Mistakes:

• Using extra data structures despite constraints
• Trying to swap data instead of nodes
• Not understanding recursion stack behavior