Reverse a Doubly Linked List in DSA C# - Execution Trace

Concept Flow - Reverse a Doubly Linked List

Start at head node

↓

Swap node.prev and node.next

↓

Move to node.prev (old next)

↓

Repeat until node is null

↓

Update head to last processed node

We start at the head, swap the previous and next pointers of each node, move forward using the old next pointer, and finally update the head to the last node processed.

Execution Sample

DSA C#

Node Reverse(Node head) {
  Node current = head;
  Node temp = null;
  while (current != null) {
    temp = current.prev;
    current.prev = current.next;
    current.next = temp;
    current = current.prev;
  }
  if (temp != null) head = temp.prev;
  return head;
}

This code reverses the doubly linked list by swapping prev and next pointers of each node and updating the head.

Execution Table

Step	Operation	Current Node	Pointer Changes	Visual State
1	Start at head (Node 1)	1	None	1 <-> 2 <-> 3 <-> null
2	Swap prev and next of Node 1	1	prev: null -> 2, next: 2 -> null	1 <- 2 3 (disconnected from 1's next)
3	Move current to old next (current.prev)	2	current moves from 1 to 2	1 <- 2 <-> 3
4	Swap prev and next of Node 2	2	prev: 1 -> 3, next: 3 -> 1	1 <- 2 <-> 3 (links reversed for 2)
5	Move current to old next (current.prev)	3	current moves from 2 to 3	1 <- 2 <- 3
6	Swap prev and next of Node 3	3	prev: 2 -> null, next: null -> 2	1 <- 2 <- 3 (3's next is null now)
7	Move current to old next (current.prev)	null	current moves from 3 to null	Traversal ends
8	Update head to last processed node (Node 3)	3	head updated from 1 to 3	3 <-> 2 <-> 1 <-> null

💡 Traversal ends when current becomes null after processing all nodes.

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	Final
current	1	2	3	null	null
temp	null	null	1	2	2
head	1	1	1	1	3

Key Moments - 3 Insights

Why do we move current to current.prev after swapping pointers?

Because after swapping, current.prev points to the original next node. This is shown in execution_table rows 3, 5, and 7 where current moves to the old next node via current.prev.

Why do we update head to temp.prev at the end?

Temp holds the previous node before current became null. Since pointers are swapped, temp.prev is the new head. This is shown in execution_table row 8 where head changes from 1 to 3.

What happens if the list is empty (head is null)?

The while loop never runs, and head remains null. The function returns null immediately, meaning the list stays empty.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of current after step 4?

A Node 2

B Node 1

C Node 3

D null

💡 Hint

Check the 'current' column in execution_table row 5.

At which step does the head pointer get updated to the new head?

A Step 6

B Step 8

C Step 7

D Step 5

💡 Hint

Look at the 'Operation' and 'Pointer Changes' columns in execution_table row 8.

If we forget to swap prev and next pointers, what would happen to the list?

A The list would be reversed correctly

B The list would become singly linked

C The list would remain unchanged

D The list would become empty

💡 Hint

Refer to the 'Pointer Changes' column in execution_table rows 2, 4, and 6 to see the effect of swapping.

Concept Snapshot

Reverse a Doubly Linked List:
- Start at head node
- For each node, swap prev and next pointers
- Move current to old next via current.prev
- Repeat until current is null
- Update head to last processed node
- Returns new head of reversed list

Full Transcript

To reverse a doubly linked list, we start at the head node. For each node, we swap its prev and next pointers. After swapping, we move to the next node using the old next pointer, which is now current.prev. We repeat this until we reach the end of the list (current becomes null). Finally, we update the head pointer to the last node processed, which becomes the new head of the reversed list. This process effectively reverses the direction of the list, so the tail becomes the new head and vice versa.

Practice

1. What is the main purpose of reversing a doubly linked list?

easy

A. To change the order of nodes so the last becomes first and vice versa

B. To delete all nodes from the list

C. To add new nodes at the end of the list

D. To sort the nodes in ascending order

Solution

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

   A doubly linked list has nodes connected forward and backward using next and prev pointers.

2. Step 2: Purpose of reversing

   Reversing swaps the direction so the last node becomes the first and the first becomes the last.

3. Final Answer:

   To change the order of nodes so the last becomes first and vice versa -> Option A

4. Quick Check:

   Reversing = changing node order [OK]

Hint: Reversing flips the list order front to back [OK]

Common Mistakes:

• Confusing reversing with deleting nodes
• Thinking reversing sorts the list
• Mixing up adding nodes with reversing

2. Which of the following C# code snippets correctly swaps the next and prev pointers of a node in a doubly linked list?

easy

A. node.next = node.prev; node.prev = node.next;

B. node.prev = node.next; node.next = node.prev;

C. var temp = node.prev; node.next = node.prev; node.prev = temp;

D. var temp = node.next; node.next = node.prev; node.prev = temp;

Solution

1. Step 1: Understand swapping pointers

   To swap two pointers, use a temporary variable to hold one value before overwriting it.

2. Step 2: Check each option

   The correct snippet saves node.next in temp, assigns node.next = node.prev, then node.prev = temp. Direct assignments without temp cause data loss. The flawed temp usage assigns node.next = node.prev without saving the original node.next first, losing data.

3. Final Answer:

   var temp = node.next; node.next = node.prev; node.prev = temp; -> Option D

4. Quick Check:

   Use temp variable to swap pointers [OK]

Hint: Always use a temp variable to swap two pointers [OK]

Common Mistakes:

• Swapping without a temporary variable
• Assigning pointers in wrong order
• Confusing next and prev roles

3. What will be the output of the following C# code after reversing the doubly linked list?

class Node {
  public int data;
  public Node next, prev;
  public Node(int d) { data = d; next = prev = null; }
}

// List: 1 <-> 2 <-> 3
Node head = new Node(1);
Node second = new Node(2);
Node third = new Node(3);
head.next = second; second.prev = head;
second.next = third; third.prev = second;

// Reverse function
Node Reverse(Node head) {
  Node current = head;
  Node temp = null;
  while (current != null) {
    temp = current.prev;
    current.prev = current.next;
    current.next = temp;
    current = current.prev;
  }
  if (temp != null) head = temp.prev;
  return head;
}

head = Reverse(head);

// Print list forward
Node tempNode = head;
string result = "";
while (tempNode != null) {
  result += tempNode.data + " -> ";
  tempNode = tempNode.next;
}
result += "null";
Console.WriteLine(result);

medium

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. null

D. 1 -> 3 -> 2 -> null

Solution

1. Step 1: Trace the reversal loop

   Each node swaps its next and prev pointers. The loop moves current to current.prev (which was originally next). After the loop, head is updated to the last processed node.

2. Step 2: Verify the new order

   Original list: 1 <-> 2 <-> 3. After reversal, the list becomes 3 <-> 2 <-> 1. Printing forward from new head prints 3 -> 2 -> 1 -> null.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option A

4. Quick Check:

   Reversed list prints backward order [OK]

Hint: Reversing flips list order; last node becomes head [OK]

Common Mistakes:

• Not updating head after reversal
• Confusing next and prev pointers
• Printing list without updating head

4. Identify the bug in this reverse function for a doubly linked list:

Node Reverse(Node head) {
  Node current = head;
  Node temp = null;
  while (current != null) {
    temp = current.next;
    current.next = current.prev;
    current.prev = temp;
    current = current.next;
  }
  if (temp != null) head = temp.prev;
  return head;
}

medium

A. The variable temp should store current.next, not current.prev

B. The loop should move current to current.prev, not current.next

C. The head update line should be head = temp.next

D. There is no bug; the code is correct

Solution

1. Step 1: Analyze pointer swapping

   The code swaps next and prev pointers correctly, but moves current to current.next, which after swapping points backward.

2. Step 2: Correct traversal direction

   After swapping, current should move to current.prev (originally next) to continue forward traversal. Moving to current.next causes skipping or infinite loop.

3. Final Answer:

   The loop should move current to current.prev, not current.next -> Option B

4. Quick Check:

   Traverse using swapped pointers correctly [OK]

Hint: After swapping, move current to current.prev pointer [OK]

Common Mistakes:

• Moving current in wrong direction after swap
• Incorrect temp assignment
• Wrong head update after reversal

5. Given a doubly linked list with nodes containing integer values, how would you modify the reversal function to also count how many nodes were reversed during the process?

hard

A. Add a counter but increment it only after the loop finishes.

B. Count nodes before reversal and return the count without changes.

C. Add a counter variable initialized to 0, increment it inside the reversal loop, and return it along with the new head.

D. Use recursion to reverse and count nodes simultaneously.

Solution

1. Step 1: Initialize a counter before the loop

   Start a counter at zero to track how many nodes are processed during reversal.

2. Step 2: Increment counter inside the reversal loop

   Each time a node is processed (pointers swapped), increase the counter by one.

3. Step 3: Return both new head and count

   Modify the function to return the new head and the count of reversed nodes.

4. Final Answer:

   Add a counter variable initialized to 0, increment it inside the reversal loop, and return it along with the new head. -> Option C

5. Quick Check:

   Count nodes during reversal loop [OK]

Hint: Count nodes inside reversal loop for accurate total [OK]

Common Mistakes:

• Counting nodes before reversal only
• Incrementing counter after loop ends
• Not returning the count with the head