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

Concept Flow - Doubly Linked List Structure and Node Design

Create Node with data

↓

Node has prev pointer

↓

Start with head pointing to first node

↓

Traverse nodes using next pointer

↓

Can move backward using prev pointer

↓

Insert/Delete nodes by updating prev and next pointers

A doubly linked list node holds data and two pointers: one to the previous node and one to the next node, allowing traversal in both directions.

Execution Sample

DSA Python

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

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	Node(10)	prev=None, next=None	None <- [10] -> None
2	Create node with data=20	Node(10), Node(20)	prev=None, next=None for new node	None <- [10] -> None None <- [20] -> None
3	Link node 10 next to node 20; node 20 prev to node 10	Node(10) <-> Node(20)	10.next=20, 20.prev=10	None <- [10] <-> [20] -> None
4	Create node with data=30	Node(10) <-> Node(20), Node(30)	prev=None, next=None for new node	None <- [10] <-> [20] -> None None <- [30] -> None
5	Link node 20 next to node 30; node 30 prev to node 20	Node(10) <-> Node(20) <-> Node(30)	20.next=30, 30.prev=20	None <- [10] <-> [20] <-> [30] -> None
6	Traverse forward from head (node 10)	Node(10) <-> Node(20) <-> Node(30)	No pointer changes	Traverse order: 10 -> 20 -> 30
7	Traverse backward from tail (node 30)	Node(10) <-> Node(20) <-> Node(30)	No pointer changes	Traverse order: 30 -> 20 -> 10
8	End of demonstration	Node(10) <-> Node(20) <-> Node(30)	No pointer changes	List stable with bidirectional links

💡 Demonstration ends after creating and linking three nodes with bidirectional pointers.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
Node(10).prev	None	None	None	None	None	None	None	None	None
Node(10).next	None	None	None	Node(20)	Node(20)	Node(20)	Node(20)	Node(20)	Node(20)
Node(20).prev	None	None	None	Node(10)	Node(10)	Node(10)	Node(10)	Node(10)	Node(10)
Node(20).next	None	None	None	None	None	Node(30)	Node(30)	Node(30)	Node(30)
Node(30).prev	None	None	None	None	None	Node(20)	Node(20)	Node(20)	Node(20)
Node(30).next	None	None	None	None	None	None	None	None	None

Key Moments - 3 Insights

Why does each node have both prev and next pointers?

Each node has prev and next pointers to allow moving forward and backward through the list, as shown in steps 3 and 5 where links are set both ways.

What does it mean when a node's prev or next pointer is None?

A None pointer means there is no node in that direction, like the first node's prev and last node's next in the visual states at steps 1 and 5.

How do we know the list is connected correctly?

By checking pointer changes in steps 3 and 5 and the visual state showing bidirectional arrows, confirming nodes link both ways.

Visual Quiz - 3 Questions

Look at the execution table at step 3. What is Node(10).next pointing to?

A Node(20)

B None

C Node(30)

D Node(10)

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns at step 3 in the execution table.

At which step does Node(30).prev get assigned to Node(20)?

A Step 2

B Step 4

C Step 5

D Step 1

💡 Hint

Look at the 'Pointer Changes' column for step 5 in the execution table.

If we remove the prev pointer from nodes, what traversal becomes impossible?

A Forward traversal

B Backward traversal

C Both forward and backward traversal

D Traversal is unaffected

💡 Hint

Refer to the concept flow and key moments explaining the role of prev pointers.

Concept Snapshot

Doubly Linked List Node:
- Holds data, prev pointer, next pointer
- prev points to previous node or None
- next points to next node or None
- Allows traversal forward and backward
- Insert/delete by updating prev and next pointers

Full Transcript

A doubly linked list node stores data and two pointers: prev and next. The prev pointer links to the previous node, and the next pointer links to the next node. This structure allows moving forward and backward through the list. Nodes at the ends have None for prev (head) or next (tail). Creating nodes and linking them updates these pointers to connect nodes bidirectionally. Traversal uses next pointers to move forward and prev pointers to move backward. This design supports flexible navigation and easy insertion or deletion by adjusting pointers.

Practice

1. What is the main difference between a doubly linked list node and a singly linked list node?

easy

A. A doubly linked list node does not store any data.

B. A doubly linked list node has pointers to both the previous and next nodes.

C. A doubly linked list node can only point to the next node.

D. A doubly linked list node stores data in a different format.

Solution

1. Step 1: Understand node pointers in singly linked list

   A singly linked list node has only one pointer to the next node.

2. Step 2: Understand node pointers in doubly linked list

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

3. Final Answer:

   A doubly linked list node has pointers to both the previous and next nodes. -> Option B

4. Quick Check:

   Doubly linked list nodes have two pointers = D [OK]

Hint: Remember: doubly means two pointers, singly means one [OK]

Common Mistakes:

• Thinking doubly linked list nodes store data differently
• Confusing pointer count between singly and doubly linked lists
• Assuming doubly linked list nodes have only one pointer

2. Which of the following is the correct way to define a doubly linked list node class in Python?

easy

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

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

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

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

Solution

1. Step 1: Check for data storage and pointers

   A doubly linked list node must store data and have both prev and next pointers.

2. Step 2: Verify class definitions

   class Node: def __init__(self, data): self.data = data self.prev = None self.next = None correctly defines data, prev, and next attributes in __init__.

3. Final Answer:

   class Node:\n def __init__(self, data):\n self.data = data\n self.prev = None\n self.next = None -> Option D

4. Quick Check:

   Node class with data, prev, next = B [OK]

Hint: Doubly linked nodes need both prev and next pointers [OK]

Common Mistakes:

• Omitting the prev pointer
• Not initializing data attribute
• Defining only next pointer like singly linked list

3. What will be the output of the following code snippet?

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

head = Node(1)
second = Node(2)
third = Node(3)

head.next = second
second.prev = head
second.next = third
third.prev = second

current = third
result = []
while current:
    result.append(current.data)
    current = current.prev
print(result)

medium

A. [3, 2, 1]

B. [1, 2, 3]

C. [1, 3, 2]

D. [3, 1, 2]

Solution

1. Step 1: Understand the linked list connections

   Nodes are linked as 1 <-> 2 <-> 3 with prev and next pointers set correctly.

2. Step 2: Trace the while loop starting from third node

   Starting at node with data 3, append 3, move to prev (2), append 2, move to prev (1), append 1, then prev is None, loop ends.

3. Final Answer:

   [3, 2, 1] -> Option A

4. Quick Check:

   Traverse backward from third node = [3, 2, 1] [OK]

Hint: Backward traversal collects nodes in reverse order [OK]

Common Mistakes:

• Confusing next and prev pointers during traversal
• Starting traversal from head instead of third
• Appending data in wrong order

4. Identify the error in the following doubly linked list node insertion code and select the correct fix.

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

head = Node(1)
new_node = Node(2)

# Insert new_node after head
new_node.next = head.next
head.next.prev = new_node
head.next = new_node
new_node.prev = head

medium

A. No error; code inserts node correctly.

B. new_node.prev should be assigned before head.next.

C. head.next.prev causes AttributeError; fix by checking if head.next is not None before accessing prev.

D. new_node.next should be assigned after head.next.

Solution

1. Step 1: Analyze head.next before insertion

   Initially, head.next is None, so head.next.prev causes AttributeError.

2. Step 2: Fix by checking if head.next exists before accessing prev

   Adding a condition to check if head.next is not None before setting head.next.prev avoids error.

3. Final Answer:

   head.next.prev causes AttributeError; fix by checking if head.next is not None before accessing prev. -> Option C

4. Quick Check:

   Check for None before accessing prev pointer [OK]

Hint: Always check if next node exists before accessing its prev [OK]

Common Mistakes:

• Ignoring None check causing runtime error
• Assigning pointers in wrong order
• Assuming head.next is always valid

5. You want to insert a new node with value 4 between nodes with values 2 and 3 in a doubly linked list. Given the nodes are connected as 1 <-> 2 <-> 3, which sequence of pointer updates correctly inserts the new node?

hard

A. new_node.prev = node2; new_node.next = node3; node2.next = new_node; node3.prev = new_node

B. new_node.next = node2; new_node.prev = node3; node3.next = new_node; node2.prev = new_node

C. node2.next = new_node; new_node.next = node3; node3.next = new_node; new_node.prev = node2

D. node3.prev = new_node; new_node.prev = node2; node2.next = new_node; new_node.next = node2

Solution

1. Step 1: Set new_node pointers to neighbors

   new_node.prev should point to node2 and new_node.next to node3 to link correctly.

2. Step 2: Update neighbors to point to new_node

   node2.next must point to new_node and node3.prev must point to new_node to complete insertion.

3. Final Answer:

   new_node.prev = node2; new_node.next = node3; node2.next = new_node; node3.prev = new_node -> Option A

4. Quick Check:

   Set new_node pointers then update neighbors [OK]

Hint: Link new node to neighbors, then neighbors back to new node [OK]

Common Mistakes:

• Swapping prev and next pointers
• Updating neighbors before setting new_node pointers
• Forgetting to update one neighbor pointer