Doubly Linked List Structure and Node Design in DSA C# - Execution Trace

Concept Flow - Doubly Linked List Structure and Node Design

Create Node with data

↓

Set node.prev = null

↓

Set node.next = null

↓

If list empty?

Yes→Set head = node, tail = node

No↓

Link node.prev to tail

↓

Link tail.next to node

↓

Update tail to new node

↓

Done

This flow shows how a new node is created with pointers and added to a doubly linked list, updating head and tail pointers accordingly.

Execution Sample

DSA C#

class Node {
  public int data;
  public Node prev;
  public Node next;

  public Node(int val) {
    data = val; prev = null; next = null;
  }
}

Defines a doubly linked list node with data and two pointers: prev and next.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Create node with data=10	None	node.prev=null, node.next=null	null <- [10] -> null
2	List empty? Yes	[10]	head=node, tail=node	[10] (head & tail)
3	Create node with data=20	[10]	node.prev=null, node.next=null	[10] (head & tail) and new node [20]
4	List empty? No	[10, 20]	node.prev=tail(10), tail.next=node(20)	[10] <-> [20]
5	Update tail to new node	[10, 20]	tail=node(20)	[10] (head) <-> [20] (tail)
6	Create node with data=30	[10, 20]	node.prev=null, node.next=null	[10] <-> [20] and new node [30]
7	List empty? No	[10, 20, 30]	node.prev=tail(20), tail.next=node(30)	[10] <-> [20] <-> [30]
8	Update tail to new node	[10, 20, 30]	tail=node(30)	[10] (head) <-> [20] <-> [30] (tail)
9	End of insertion	[10, 20, 30]	No changes	[10] (head) <-> [20] <-> [30] (tail)

💡 All nodes inserted, list updated with correct prev and next pointers.

Variable Tracker

Variable	Start	After Adding 10	After Adding 20	After Adding 30	Final
node.data	N/A	10	20	30	30
node.prev	N/A	null	tail(10)	tail(20)	tail(20)
node.next	N/A	null	null	null	null
head	null	node(10)	node(10)	node(10)	node(10)
tail	null	node(10)	node(20)	node(30)	node(30)
List Size	0	1	2	3	3

Key Moments - 3 Insights

Why do we set node.prev to tail before linking tail.next to node?

Setting node.prev to tail first ensures the new node correctly points back to the last node before tail.next points forward to the new node. See execution_table rows 4 and 7 where node.prev is set before tail.next.

What happens if the list is empty when inserting the first node?

If the list is empty, both head and tail point to the new node, as shown in execution_table row 2. This initializes the list with one node.

Why do we update tail to the new node after linking?

Updating tail to the new node marks it as the last node in the list, so future insertions link correctly. This is shown in rows 5 and 8.

Visual Quiz - 3 Questions

Look at the execution_table, what is the tail node data after step 5?

A 10

B 30

C 20

D null

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns at step 5.

At which step does the list first have two nodes linked?

A Step 2

B Step 4

C Step 3

D Step 6

💡 Hint

Look for when node.prev and tail.next pointers are set to link nodes.

If we skip setting node.prev before tail.next, what would happen?

A Nodes would not link backward correctly

B List would be empty

C Tail pointer would be null

D Head pointer would change

💡 Hint

Refer to key_moments about pointer linking order.

Concept Snapshot

Doubly Linked List Node:
- Contains data, prev pointer, next pointer
- prev points to previous node or null
- next points to next node or null
Insertion:
- If empty, head and tail = new node
- Else, link new node.prev to tail
- Link tail.next to new node
- Update tail to new node

Full Transcript

A doubly linked list node holds data and two pointers: prev and next. When inserting a node, we create it with null pointers. If the list is empty, head and tail point to this node. Otherwise, we link the new node's prev to the current tail, then link the tail's next to the new node, and finally update tail to the new node. This maintains a chain where each node points forward and backward, allowing easy traversal in both directions.

Practice

1. What is the main advantage of a Doubly Linked List compared to a singly linked list?

easy

A. It allows traversal in both forward and backward directions.

B. It uses less memory than a singly linked list.

C. It does not require a head pointer.

D. It automatically sorts the elements.

Solution

1. Step 1: Understand the structure difference

   A doubly linked list has nodes with two pointers: one to the next node and one to the previous node.

2. Step 2: Identify traversal capabilities

   This allows moving forward and backward through the list, unlike singly linked lists which only move forward.

3. Final Answer:

   It allows traversal in both forward and backward directions. -> Option A

4. Quick Check:

   Doubly linked list = two-way traversal [OK]

Hint: Remember: two pointers mean two directions [OK]

Common Mistakes:

• Thinking it uses less memory than singly linked list
• Assuming it doesn't need a head pointer
• Believing it sorts elements automatically

2. Which of the following is the correct way to declare a node class for a doubly linked list in C#?

easy

A. class Node { public int data; public Node next; }

B. public class Node { public int data; public Node next; public Node prev; }

C. public class Node { int data; Node next; }

D. public struct Node { int data; Node next; Node prev; }

Solution

1. Step 1: Check for public access and all pointers

   The node must have public fields for data, next, and prev to represent doubly linked list nodes.

2. Step 2: Verify syntax correctness

   public class Node { public int data; public Node next; public Node prev; } declares all required fields with public access, suitable for a doubly linked list node.

3. Final Answer:

   public class Node { public int data; public Node next; public Node prev; } -> Option B

4. Quick Check:

   Node class with data, next, prev = public class Node { public int data; public Node next; public Node prev; } [OK]

Hint: Doubly linked node needs next and prev pointers public [OK]

Common Mistakes:

• Omitting the prev pointer
• Using struct instead of class (less common for linked lists)
• Not making fields public for access

3. What will be the output of the following C# code snippet for a doubly linked list node?

Node node = new Node();
node.data = 10;
node.next = null;
node.prev = null;
Console.WriteLine(node.data);

medium

A. 10

B. null

C. 0

D. Compilation error

Solution

1. Step 1: Analyze node initialization

   The node's data is set to 10 explicitly, and next and prev are set to null.

2. Step 2: Check the printed value

   The Console.WriteLine prints node.data, which is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   node.data = 10 prints 10 [OK]

Hint: Prints the assigned data value, not pointers [OK]

Common Mistakes:

• Confusing null pointers with data value
• Expecting default 0 instead of assigned 10
• Thinking it causes compilation error

4. Identify the error in this doubly linked list node insertion code snippet:

Node newNode = new Node();
newNode.data = 5;
newNode.next = head;
head.prev = newNode;
head = newNode;

medium

A. Missing initialization of newNode.prev to null

B. head.prev should be assigned after head is updated

C. No error, code correctly inserts at the beginning

D. newNode.next should be null

Solution

1. Step 1: Understand insertion at list start

   newNode's next points to current head, head's prev points back to newNode, then head updates to newNode.

2. Step 2: Check pointer assignments

   newNode.prev is null by default or not needed explicitly here; the code correctly links nodes.

3. Final Answer:

   No error, code correctly inserts at the beginning -> Option C

4. Quick Check:

   Insertion at head with correct pointers = No error, code correctly inserts at the beginning [OK]

Hint: Check pointer updates order carefully [OK]

Common Mistakes:

• Assuming prev must be set explicitly to null
• Thinking head.prev must be set after head update
• Confusing next pointer assignment

5. You want to design a method to remove a node from the middle of a doubly linked list. Which pointers must be updated to maintain list integrity?

hard

A. Only the head pointer

B. The tail pointer and the node's prev pointer

C. Only the node's next pointer

D. The previous node's next and the next node's prev pointers

Solution

1. Step 1: Identify nodes around the removed node

   To remove a node, you must link its previous node to its next node directly.

2. Step 2: Update pointers to bypass the removed node

   Set previous node's next to removed node's next, and next node's prev to removed node's prev.

3. Final Answer:

   The previous node's next and the next node's prev pointers -> Option D

4. Quick Check:

   Update prev.next and next.prev to remove node [OK]

Hint: Update both prev and next pointers around removed node [OK]

Common Mistakes:

• Only updating one pointer and breaking the list
• Trying to update head or tail unnecessarily
• Ignoring the prev pointer of the next node