Creating a Singly Linked List from Scratch in DSA Java - Visual Walkthrough

Concept Flow - Creating a Singly Linked List from Scratch

Create new node with data

↓

Is list empty? head == null

↓

Set head = new node

↓

Set last node's next = new node

↓

Done

Start by making a new node. If the list is empty, set head to this node. Otherwise, go to the last node and link it to the new node.

Execution Sample

DSA Java

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

Node head = null;
head = new Node(10);

This code creates a node with data 10 and sets it as the head of the list.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Create new node with data=10	None	None	null
2	Check if head == null	None	head == null (true)	null
3	Set head = new node	10	head -> Node(10)	10 -> null
4	Create new node with data=20	10	None	10 -> null
5	Check if head == null	10	head != null (false)	10 -> null
6	Traverse to last node (Node with data=10)	10	current -> Node(10)	10 -> null
7	Set last node's next = new node	10 -> 20	Node(10).next -> Node(20)	10 -> 20 -> null
8	Create new node with data=30	10 -> 20	None	10 -> 20 -> null
9	Check if head == null	10 -> 20	head != null (false)	10 -> 20 -> null
10	Traverse to last node (Node with data=20)	10 -> 20	current -> Node(20)	10 -> 20 -> null
11	Set last node's next = new node	10 -> 20 -> 30	Node(20).next -> Node(30)	10 -> 20 -> 30 -> null
12	End of insertions	10 -> 20 -> 30	No changes	10 -> 20 -> 30 -> null

💡 All nodes inserted; list is complete with 3 nodes linked.

Variable Tracker

Variable	Start	After Step 3	After Step 7	After Step 11	Final
head	null	Node(10)	Node(10)	Node(10)	Node(10)
current	null	null	Node(10)	Node(20)	Node(20)
new node	null	Node(10)	Node(20)	Node(30)	Node(30)
List Size	0	1	2	3	3

Key Moments - 3 Insights

Why do we check if head is null before inserting a new node?

Because if head is null, the list is empty and the new node becomes the first node (head). This is shown in execution_table row 2 and 3.

Why do we traverse to the last node before inserting a new node when the list is not empty?

Because we need to add the new node at the end. Traversing finds the last node whose next is null, as shown in rows 6 and 10.

What does setting last node's next to new node do?

It links the new node to the end of the list, extending it. This pointer change is shown in rows 7 and 11.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the visual state of the list?

A null

B 20 -> null

C 10 -> null

D 10 -> 20 -> null

💡 Hint

Check the Visual State column at step 3 in execution_table.

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

A Step 7

B Step 5

C Step 10

D Step 11

💡 Hint

Look at the Nodes in List and Visual State columns to see when two nodes appear.

If we skip the traversal step before inserting a new node when list is not empty, what happens?

A New node becomes head

B New node is not linked to list

C List size decreases

D Last node is updated automatically

💡 Hint

Refer to key_moments about why traversal is needed before linking new node.

Concept Snapshot

Creating a singly linked list:
- Create a node with data and next=null
- If list empty (head == null), set head = new node
- Else, traverse to last node (next == null)
- Set last node's next = new node
- Repeat to add more nodes
- List ends with last node's next = null

Full Transcript

To create a singly linked list from scratch, we start by making a new node with the data we want. If the list is empty, meaning head is null, we set head to this new node. If the list already has nodes, we move through the list from head until we find the last node, which has next as null. Then we link this last node's next pointer to the new node. This process repeats for each new node we want to add. The list always ends with a node whose next pointer is null, showing the end of the list.

Practice

1. What is the main purpose of a Singly Linked List in data structures?

easy

A. To store elements in a sequence where each element points to the next

B. To store elements with keys and values for fast lookup

C. To store elements in a sorted array for quick access

D. To store elements in a tree structure with parent and child nodes

Solution

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

   A singly linked list stores elements in nodes where each node points to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike arrays or trees, singly linked lists do not store keys or have parent-child relationships; they only link sequentially.

3. Final Answer:

   To store elements in a sequence where each element points to the next -> Option A

4. Quick Check:

   Singly Linked List = sequence with next pointers [OK]

Hint: Remember: singly linked list nodes link one after another [OK]

Common Mistakes:

• Confusing linked list with arrays or hash maps
• Thinking linked list stores key-value pairs
• Assuming linked list nodes have multiple pointers

2. Which of the following is the correct way to declare a node class for a singly linked list in Java?

easy

A. class Node { int data; Node left; Node right; }

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

C. class Node { int data; Node prev; }

D. class Node { int data; int next; }

Solution

1. Step 1: Identify the node structure for singly linked list

   A singly linked list node contains data and a reference to the next node only.

2. Step 2: Check each option

   class Node { int data; Node next; } has data and next pointer; class Node { int data; Node prev; } has prev (used in doubly linked list); class Node { int data; Node left; Node right; } has left and right (used in trees); class Node { int data; int next; } incorrectly uses int for next pointer.

3. Final Answer:

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

4. Quick Check:

   Node = data + next pointer [OK]

Hint: Node must have 'next' pointer, not 'prev' or others [OK]

Common Mistakes:

• Using 'prev' pointer instead of 'next'
• Using primitive type for next pointer
• Confusing tree node with linked list node

3. What will be the output of the following code snippet that creates a singly linked list with nodes 10, 20, 30 and prints the list?

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

public class LinkedList {
  Node head;
  public void add(int data) {
    Node newNode = new Node(data);
    if (head == null) {
      head = newNode;
    } else {
      Node temp = head;
      while (temp.next != null) {
        temp = temp.next;
      }
      temp.next = newNode;
    }
  }
  public void printList() {
    Node temp = head;
    while (temp != null) {
      System.out.print(temp.data + " -> ");
      temp = temp.next;
    }
    System.out.print("null");
  }

  public static void main(String[] args) {
    LinkedList list = new LinkedList();
    list.add(10);
    list.add(20);
    list.add(30);
    list.printList();
  }
}

medium

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

B. null

C. 10 -> 20 -> null

D. 10 -> 20 -> 30 -> null

Solution

1. Step 1: Understand the add method behavior

   The add method appends new nodes at the end of the list, preserving insertion order.

2. Step 2: Trace the printList method

   It prints nodes from head to last, showing 10, then 20, then 30, followed by null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option D

4. Quick Check:

   Adding nodes appends at end, print shows 10->20->30->null [OK]

Hint: Add appends at end; print shows order added [OK]

Common Mistakes:

• Assuming add inserts at head instead of tail
• Printing nodes in reverse order
• Missing the final 'null' in output

4. Identify the error in the following code that attempts to add a node to the singly linked list:

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

public class LinkedList {
  Node head;
  public void add(int data) {
    Node newNode = new Node(data);
    if (head == null) {
      head = newNode;
    } else {
      Node temp = head;
      while (temp != null) {
        temp = temp.next;
      }
      temp.next = newNode;
    }
  }
}

medium

A. The while loop condition should be 'temp.next != null' instead of 'temp != null'

B. The newNode should be created outside the add method

C. The head should be initialized to newNode inside the else block

D. The Node class should have a 'prev' pointer

Solution

1. Step 1: Analyze the while loop condition

   The loop currently runs while temp != null, which moves temp to null and loses reference to last node.

2. Step 2: Correct the loop to stop at last node

   The loop should run while temp.next != null to stop at the last node, so we can set temp.next = newNode.

3. Final Answer:

   The while loop condition should be 'temp.next != null' instead of 'temp != null' -> Option A

4. Quick Check:

   Loop stops at last node using temp.next != null [OK]

Hint: Loop to last node uses 'temp.next != null' [OK]

Common Mistakes:

• Looping until temp is null loses last node reference
• Creating newNode outside add method is unnecessary
• Adding 'prev' pointer in singly linked list node

5. You want to create a singly linked list from scratch that supports adding nodes at the head and printing the list. Which code snippet correctly implements adding at the head?

hard

A. public void addAtHead(int data) { Node newNode = new Node(data); head = newNode; newNode.next = head; }

B. public void addAtHead(int data) { Node newNode = new Node(data); head.next = newNode; head = newNode; }

C. public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; }

D. public void addAtHead(int data) { Node newNode = new Node(data); newNode = head; head = newNode; }

Solution

1. Step 1: Understand adding at head logic

   To add at head, new node's next should point to current head, then head updated to new node.

2. Step 2: Check each option's correctness

   public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; } correctly sets newNode.next = head and then head = newNode. Others incorrectly assign or reverse pointers.

3. Final Answer:

   public void addAtHead(int data) { Node newNode = new Node(data); newNode.next = head; head = newNode; } -> Option C

4. Quick Check:

   Add at head = newNode.next = head; head = newNode [OK]

Hint: New head points to old head, then update head [OK]

Common Mistakes:

• Assigning head.next before updating head
• Reversing pointer assignments
• Overwriting newNode reference incorrectly