What if you could jump back and forth in your task list without any hassle or mistakes?
Why Create and Initialize Doubly Linked List in DSA Javascript?
Imagine you have a list of tasks written on sticky notes. You want to add new tasks, remove old ones, and sometimes look at the previous or next task easily.
If you try to manage this list by just writing tasks on a single strip of paper, it becomes hard to jump back and forth or insert tasks in the middle.
Using a simple list or array means you can only move forward easily. Going backward or inserting tasks in the middle means shifting many tasks manually, which is slow and error-prone.
You might lose track of tasks or accidentally overwrite them.
A doubly linked list is like having sticky notes connected with strings on both sides. Each task knows its previous and next task, so you can move forward or backward easily.
Adding or removing tasks anywhere becomes quick and safe without shifting everything.
let tasks = ['Task1', 'Task2', 'Task3']; tasks.splice(1, 0, 'NewTask'); // Insert in middle // Need to shift elements manually
class Node {
constructor(value) {
this.value = value;
this.next = null;
this.prev = null;
}
}
// Insert new node between nodes easily without shiftingIt enables smooth navigation and quick updates in lists where you need to move both forward and backward.
Music players use doubly linked lists to move to the next or previous song quickly without restarting the list.
Doubly linked lists let you move forward and backward easily.
They make inserting and deleting tasks fast and safe.
They solve the problem of shifting elements in simple lists.
Practice
Solution
Step 1: Understand node structure in doubly linked list
Each node in a doubly linked list has two pointers: one to the next node and one to the previous node.Step 2: Compare with singly linked list
Singly linked list nodes have only one pointer to the next node, so they cannot go backward.Final Answer:
A doubly linked list has nodes with pointers to both previous and next nodes. -> Option AQuick Check:
Doubly linked list = two pointers per node [OK]
- Thinking doubly linked lists use arrays
- Believing doubly linked lists only go forward
- Confusing pointers with data storage
Solution
Step 1: Identify required pointers for doubly linked list node
A doubly linked list node needs both 'prev' and 'next' pointers initialized to null.Step 2: Check constructor parameters and properties
class Node { constructor(data) { this.data = data; this.prev = null; this.next = null; } } correctly defines 'data', 'prev', and 'next' properties in the constructor.Final Answer:
class Node { constructor(data) { this.data = data; this.prev = null; this.next = null; } } -> Option BQuick Check:
Node needs prev and next pointers [OK]
- Omitting the prev pointer
- Using wrong pointer names like left/right
- Using different property names for data
class Node {
constructor(data) {
this.data = data;
this.prev = null;
this.next = null;
}
}
class DoublyLinkedList {
constructor() {
this.head = null;
this.tail = null;
}
append(data) {
const newNode = new Node(data);
if (!this.head) {
this.head = newNode;
this.tail = newNode;
} else {
this.tail.next = newNode;
newNode.prev = this.tail;
this.tail = newNode;
}
}
}
const list = new DoublyLinkedList();
list.append(1);
list.append(2);
list.append(3);
let current = list.head;
const result = [];
while (current) {
result.push(current.data);
current = current.next;
}
console.log(result);Solution
Step 1: Append nodes 1, 2, 3 to the list
Each append adds a new node at the tail, linking prev and next pointers correctly.Step 2: Traverse from head to tail collecting data
Starting at head (1), moving next collects 1, then 2, then 3, then stops at null.Final Answer:
[1, 2, 3] -> Option DQuick Check:
Traversal from head collects [1, 2, 3] [OK]
- Confusing traversal direction
- Not updating tail pointer on append
- Pushing null values instead of data
class Node {
constructor(data) {
this.data = data;
this.prev = null;
this.next = null;
}
}
class DoublyLinkedList {
constructor() {
this.head = null;
this.tail = null;
}
append(data) {
const newNode = new Node(data);
if (!this.head) {
this.head = newNode;
this.tail = newNode;
} else {
this.tail.next = newNode;
this.tail.prev = newNode;
this.tail = newNode;
}
}
}Solution
Step 1: Analyze pointer updates in append method
When adding newNode, the previous tail's next should point to newNode, and newNode's prev should point to old tail.Step 2: Identify incorrect pointer assignment
The code incorrectly sets 'this.tail.prev = newNode;' which wrongly changes the old tail's prev pointer instead of newNode's prev pointer.Final Answer:
The line 'this.tail.prev = newNode;' should be 'newNode.prev = this.tail;'. -> Option CQuick Check:
Correct prev pointer must link newNode to old tail [OK]
- Swapping prev and next pointers
- Not updating newNode.prev correctly
- Forgetting to update tail pointer
Solution
Step 1: Append values in order to the list
Using a for-of loop appends 10, then 20, then 30, building the list head to tail.Step 2: Traverse backward from tail using prev pointer
Starting at tail (30), moving prev prints 30, 20, 10 in reverse order.Final Answer:
const list = new DoublyLinkedList(); for (const val of [10, 20, 30]) list.append(val); let current = list.tail; while (current) { console.log(current.data); current = current.prev; } -> Option AQuick Check:
Backward traversal uses tail and prev pointer [OK]
- Traversing backward from head
- Using next pointer to go backward
- Appending values in reverse order but traversing forward
Ready to prove it?
Pass the test and this pattern moves unknown
