Delete Node at Beginning in DSA Go - Execution Trace

Concept Flow - Delete Node at Beginning

Start with head node

↓

Check if list is empty?

↓

Return nil

↓

Save head's next node

↓

Set head to next node

↓

Return updated head

This flow shows how to remove the first node by moving the head pointer to the next node, or returning nil if the list is empty.

Execution Sample

DSA Go

func deleteAtBeginning(head *Node) *Node {
    if head == nil {
        return nil
    }
    return head.Next
}

Deletes the first node of a singly linked list by returning the second node as the new head.

Execution Table

Step	head	Condition head == nil?	Action	New head	Output List
1	1 -> 2 -> 3 -> nil	False	Save head.Next (2 -> 3 -> nil)	2 -> 3 -> nil	1 -> 2 -> 3 -> nil (before deletion)
2	2 -> 3 -> nil	N/A	Return new head	2 -> 3 -> nil	2 -> 3 -> nil (after deletion)
3	nil	True	Return nil	nil	Empty list (nil)

💡 Execution stops when head is nil (empty list) or after returning the new head.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
head	1 -> 2 -> 3 -> nil	2 -> 3 -> nil	2 -> 3 -> nil	2 -> 3 -> nil
newHead	N/A	2 -> 3 -> nil	2 -> 3 -> nil	2 -> 3 -> nil

Key Moments - 2 Insights

Why do we check if head is nil before deleting?

Because if the list is empty (head is nil), there is no node to delete. The execution_table row 3 shows this check and returns nil immediately.

Does deleting the first node remove it from memory?

In Go, returning head.Next disconnects the first node from the list, so it can be garbage collected if no other references exist. The execution_table row 2 shows the new head after deletion.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of head after step 1?

A 1 -> 2 -> 3 -> nil

B 2 -> 3 -> nil

C nil

D 3 -> nil

💡 Hint

Check the 'head' column in row 1 of execution_table.

At which step does the function return nil?

A Step 1

B Step 2

C Step 3

D Never

💡 Hint

Look at the 'Condition head == nil?' column and the 'Action' in execution_table row 3.

If the list was initially nil, what would be the output list after deletion?

A nil

B 1 -> 2 -> 3 -> nil

C 2 -> 3 -> nil

D Empty list with one node

💡 Hint

Refer to execution_table row 3 where head is nil at start.

Concept Snapshot

Delete Node at Beginning:
- Check if list is empty (head == nil).
- If empty, return nil.
- Else, move head to head.Next.
- New head is the list without first node.
- Old first node is removed from list.

Full Transcript

To delete the first node in a singly linked list, we first check if the list is empty by seeing if head is nil. If it is nil, we return nil immediately because there is no node to delete. If the list is not empty, we move the head pointer to the next node (head.Next). This effectively removes the first node from the list. The old first node will be removed from the list and can be garbage collected if no other references exist. The function then returns the new head of the list. This process is simple and efficient, requiring only a pointer update.

Practice

1. What happens when you delete the first node in a singly linked list?

easy

A. The head pointer moves to the second node

B. The last node is deleted

C. The list becomes empty regardless of nodes

D. The middle node is deleted

Solution

1. Step 1: Understand the head pointer role

   The head pointer points to the first node in the list.

2. Step 2: Effect of deleting the first node

   Deleting the first node means moving the head pointer to the next node, which becomes the new first node.

3. Final Answer:

   The head pointer moves to the second node -> Option A

4. Quick Check:

   Deleting first node updates head = next node [OK]

Hint: Deleting first node means head points to next node [OK]

Common Mistakes:

• Thinking last node is deleted
• Assuming list becomes empty always
• Deleting middle node by mistake

2. Which of the following Go code snippets correctly deletes the first node of a singly linked list with head pointer head?

easy

A. head.next = nil

B. head.next = head

C. head = nil.next

D. head = head.next

Solution

1. Step 1: Understand pointer update to delete first node

   To delete the first node, head must point to the second node, which is head.next.

2. Step 2: Check each option for correctness

   head = head.next correctly assigns head = head.next. Others either assign incorrectly or cause errors.

3. Final Answer:

   head = head.next -> Option D

4. Quick Check:

   head = head.next deletes first node [OK]

Hint: Assign head to head.next to delete first node [OK]

Common Mistakes:

• Assigning head.next to head (wrong direction)
• Using nil.next which causes error
• Setting head.next to nil deletes wrong node

3. Given the following Go code for a singly linked list, what will be printed after deleting the first node?

package main
import "fmt"

type Node struct {
    value int
    next  *Node
}

func main() {
    head := &Node{value: 1}
    head.next = &Node{value: 2}
    head.next.next = &Node{value: 3}

    // Delete first node
    head = head.next

    // Print list
    current := head
    for current != nil {
        fmt.Print(current.value, " -> ")
        current = current.next
    }
    fmt.Print("null")
}

medium

A. 2 -> 3 -> null

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

C. 3 -> null

D. null

Solution

1. Step 1: Initial list nodes and values

   The list starts as 1 -> 2 -> 3 -> null.

2. Step 2: Deleting first node by head = head.next

   Head moves from node with value 1 to node with value 2, so list becomes 2 -> 3 -> null.

3. Final Answer:

   2 -> 3 -> null -> Option A

4. Quick Check:

   After deletion, list starts at second node [OK]

Hint: After deletion, list starts from second node [OK]

Common Mistakes:

• Printing original list without deletion
• Assuming first node still prints
• Confusing null with empty list

4. The following Go code attempts to delete the first node of a singly linked list but causes a runtime error. What is the error and how to fix it?

package main

type Node struct {
    value int
    next  *Node
}

func main() {
    var head *Node = nil

    // Delete first node
    head = head.next
}

medium

A. Compile error: cannot assign head.next to head

B. Runtime panic: nil pointer dereference; fix by checking if head is nil before deletion

C. No error; code works fine

D. Syntax error: missing import; fix by adding fmt package

Solution

1. Step 1: Identify cause of runtime panic

   head is nil, so accessing head.next causes nil pointer dereference panic.

2. Step 2: Fix by checking if head is nil before deleting

   Check if head != nil before assigning head = head.next to avoid panic.

3. Final Answer:

   Runtime panic: nil pointer dereference; fix by checking if head is nil before deletion -> Option B

4. Quick Check:

   Nil check prevents runtime panic [OK]

Hint: Always check head != nil before deleting first node [OK]

Common Mistakes:

• Ignoring nil check causing panic
• Assuming head is never nil
• Confusing syntax and runtime errors

5. You have a singly linked list with nodes: 10 -> 20 -> 30 -> 40 -> null. Write a Go function to delete the first node and return the new head. What will be the list after deletion?

func deleteFirstNode(head *Node) *Node {
    if head == nil {
        return nil
    }
    head = head.next
    return head
}

hard

A. 30 -> 40 -> null

B. 10 -> 20 -> 30 -> 40 -> null

C. 20 -> 30 -> 40 -> null

D. null

Solution

1. Step 1: Understand function logic

   If head is nil, return nil. Otherwise, move head to head.next and return it.

2. Step 2: Apply function to list 10 -> 20 -> 30 -> 40 -> null

   Deleting first node (10) moves head to node with value 20, so list becomes 20 -> 30 -> 40 -> null.

3. Final Answer:

   20 -> 30 -> 40 -> null -> Option C

4. Quick Check:

   Function returns list starting from second node [OK]

Hint: Return head.next to delete first node and update list [OK]

Common Mistakes:

• Returning original head without deletion
• Not handling nil head case
• Deleting wrong node by mistake