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

Concept Flow - Creating a Singly Linked List from Scratch

Start: Empty List

↓

Create new node with data

↓

Is list empty?

No→Traverse to last node

|Yes↓

Set head to new node

↓

Done: List updated

We start with an empty list, create a new node, check if the list is empty, if yes set head to new node, else traverse to last node and link new node there.

Execution Sample

DSA Go

type Node struct {
  data int
  next *Node
}

var head *Node = nil

// Create new node with data 10
newNode := &Node{data: 10, next: nil}

// If list empty, head = newNode

This code creates a singly linked list node with data 10 and sets it as head if the list is empty.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Start with empty list	None	head = nil	null
2	Create new node with data=10	Node(10)	newNode points to Node(10)	null
3	Check if list empty (head == nil)	Node(10)	head == nil is true	null
4	Set head = newNode	Node(10)	head -> Node(10)	head -> 10 -> null
5	Done	Node(10)	head -> Node(10)	head -> 10 -> null

💡 List created with one node; head points to Node(10), no next node.

Variable Tracker

Variable	Start	After Step 2	After Step 4	Final
head	nil	nil	Node(10)	Node(10)
newNode	nil	Node(10)	Node(10)	Node(10)

Key Moments - 2 Insights

Why do we check if head is nil before adding the new node?

Because if head is nil, the list is empty and the new node becomes the first node (head). This is shown in execution_table step 3 and 4.

What does the visual state 'head -> 10 -> null' mean?

It means the head pointer points to a node containing data 10, and this node's next pointer is null, indicating the end of the list. See execution_table step 4.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the condition checked?

A newNode == nil

B head.next == nil

C head == nil

D newNode.next == nil

💡 Hint

Check the 'Operation' column at step 3 in execution_table.

At which step does head start pointing to the new node?

A Step 4

B Step 3

C Step 2

D Step 5

💡 Hint

Look at 'Pointer Changes' column to see when head changes from nil.

If the list was not empty, what would be the next step after creating newNode?

A Set head to newNode

B Traverse to last node and link newNode

C Delete newNode

D Set newNode.next to head

💡 Hint

Refer to concept_flow ASCII diagram for the 'No' branch after 'Is list empty?'

Concept Snapshot

Creating a singly linked list:
- Start with head = nil (empty list)
- Create a new node with data and next = nil
- If head is nil, set head = new node
- Else, traverse to last node and link new node
- New node becomes the last node with next = nil

Full Transcript

We start with an empty singly linked list where head is nil. We create a new node with the given data and next pointer set to nil. We check if the list is empty by seeing if head is nil. If it is empty, we set head to point to the new node. This makes the new node the first and only node in the list. The visual state after this shows head pointing to the node with data 10 and next as nil, indicating the end of the list. If the list was not empty, we would traverse to the last node and link the new node there. This process builds the singly linked list from scratch step by step.

Practice

1. What is the main purpose of a singly linked list in data structures?

easy

A. To store elements in a hash table for fast lookup

B. To store elements in a tree-like structure with multiple branches

C. To store elements in a fixed-size array for quick access

D. To store elements in a linear order where each element points to the next

Solution

1. Step 1: Understand singly linked list structure

   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 trees or arrays, singly linked lists have a linear sequence with pointers only to the next element.

3. Final Answer:

   To store elements in a linear order where each element points to the next -> Option D

4. Quick Check:

   Singly linked list = linear chain with next pointers [OK]

Hint: Remember: singly linked list nodes point only forward [OK]

Common Mistakes:

• Confusing singly linked list with trees or arrays
• Thinking nodes have pointers to previous nodes
• Assuming fixed size storage like arrays

2. Which of the following is the correct way to define a node struct for a singly linked list in Go?

easy

A. type Node struct { value int; next *Node }

B. type Node struct { value int; prev *Node }

C. type Node struct { value int; next Node }

D. type Node struct { value int; next interface{} }

Solution

1. Step 1: Identify correct pointer usage

   In Go, the next node must be a pointer to Node, so *Node is correct.

2. Step 2: Check struct fields

   Field 'value' stores data, 'next' points to the next node. 'prev' is for doubly linked lists, and 'interface{}' is too generic.

3. Final Answer:

   type Node struct { value int; next *Node } -> Option A

4. Quick Check:

   Node next pointer = *Node [OK]

Hint: Use *Node for next pointer in Go struct [OK]

Common Mistakes:

• Using value type instead of pointer for next
• Confusing singly linked list with doubly linked list (prev pointer)
• Using interface{} instead of *Node for next

3. What will be the output of the following Go code that creates a singly linked list with nodes 1 -> 2 -> 3?

package main
import "fmt"
type Node struct {
  value int
  next  *Node
}
func main() {
  head := &Node{value: 1}
  second := &Node{value: 2}
  third := &Node{value: 3}
  head.next = second
  second.next = third
  current := head
  for current != nil {
    fmt.Print(current.value)
    if current.next != nil {
      fmt.Print(" -> ")
    }
    current = current.next
  }
  fmt.Println(" -> null")
}

medium

A. 1 -> 2 -> null

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

C. 1 -> 2 -> 3 -> null

D. Compilation error

Solution

1. Step 1: Trace node creation and linking

   Nodes with values 1, 2, 3 are created and linked in order: head.next = second, second.next = third.

2. Step 2: Trace printing loop

   Loop prints current.value, then ' -> ' if next exists, moving from head to third, then prints ' -> null'.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option C

4. Quick Check:

   Print nodes in order 1,2,3 with arrows [OK]

Hint: Follow next pointers from head to end [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing last node in output
• Confusing pointer assignments causing nil errors

4. Identify the error in this Go code snippet that tries to append a new node with value 4 to the end of a singly linked list:

func appendNode(head *Node, val int) {
  newNode := &Node{value: val}
  current := head
  for current.next != nil {
    current = current.next
  }
  current = newNode
}

medium

A. The loop condition should be current != nil instead of current.next != nil

B. The new node is not linked to the list; current pointer is reassigned incorrectly

C. The new node should be created with next pointing to head

D. The function should return the new head pointer

Solution

1. Step 1: Analyze the loop and pointer update

   The loop correctly finds the last node where current.next is nil.

2. Step 2: Check how new node is linked

   Assigning current = newNode only changes the local variable, not the last node's next pointer. It should be current.next = newNode.

3. Final Answer:

   The new node is not linked to the list; current pointer is reassigned incorrectly -> Option B

4. Quick Check:

   Link new node by current.next = newNode [OK]

Hint: Assign new node to current.next, not current [OK]

Common Mistakes:

• Reassigning current instead of current.next
• Wrong loop condition causing nil pointer dereference
• Not returning updated head when needed

5. You want to create a function in Go that reverses a singly linked list. Which approach correctly reverses the list by changing the next pointers?

func reverseList(head *Node) *Node {
  var prev *Node = nil
  current := head
  for current != nil {
    nextTemp := current.next
    current.next = prev
    prev = current
    current = nextTemp
  }
  return prev
}

What will be the output if the original list is 1 -> 2 -> 3 -> null and you print the reversed list?

hard

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

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

C. null

D. Runtime error due to nil pointer

Solution

1. Step 1: Understand reversal logic

   The function iterates through the list, reversing each node's next pointer to point to the previous node.

2. Step 2: Trace reversal on list 1 -> 2 -> 3

   After reversal, the list becomes 3 -> 2 -> 1 -> null, and prev points to new head.

3. Final Answer:

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

4. Quick Check:

   Reversed list head = prev = 3 [OK]

Hint: Reverse pointers by tracking prev, current, next [OK]

Common Mistakes:

• Not updating current.next correctly
• Losing track of next node causing infinite loop
• Returning old head instead of new head