Delete Node by Value in DSA Go

Mental Model

To remove a node with a specific value, find it first, then change the previous node to skip it.

Analogy: Imagine a chain of paper clips linked together. To remove one clip, you open the clip before it and link it directly to the clip after the one you want to remove.

head -> 1 -> 2 -> 3 -> 4 -> null

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> 4, delete value 3

Goal: Remove the node containing value 3 from the list

Step 1: start at head node with value 1

head -> [curr->1] -> 2 -> 3 -> 4 -> null

Why: We begin searching from the first node

Step 2: move curr to node with value 2

head -> 1 -> [curr->2] -> 3 -> 4 -> null

Why: Keep moving forward to find the node before the one to delete

Step 3: curr.next value is 3, which matches target; remove node 3 by linking curr.next to node 4

head -> 1 -> 2 -> 4 -> null

Why: Skip the node with value 3 to delete it

Result:

head -> 1 -> 2 -> 4 -> null

Annotated Code

DSA Go

package main

import "fmt"

type Node struct {
	value int
	next  *Node
}

// deleteNodeByValue removes the first node with the given value from the list
func deleteNodeByValue(head *Node, val int) *Node {
	// handle empty list
	if head == nil {
		return head
	}

	// if head needs to be deleted
	if head.value == val {
		return head.next
	}

	curr := head
	// traverse list to find node before the one to delete
	for curr.next != nil && curr.next.value != val {
		curr = curr.next
	}

	// if found, skip the node
	if curr.next != nil {
		curr.next = curr.next.next
	}

	return head
}

// printList prints the linked list values
func printList(head *Node) {
	curr := head
	for curr != nil {
		fmt.Printf("%d", curr.value)
		if curr.next != nil {
			fmt.Printf(" -> ")
		}
		curr = curr.next
	}
	fmt.Println(" -> null")
}

func main() {
	head := &Node{value: 1}
	head.next = &Node{value: 2}
	head.next.next = &Node{value: 3}
	head.next.next.next = &Node{value: 4}

	fmt.Print("Original list: ")
	printList(head)

	head = deleteNodeByValue(head, 3)

	fmt.Print("After deleting 3: ")
	printList(head)
}

if head == nil { return head }

handle empty list case early

if head.value == val { return head.next }

if head is the node to delete, remove it by returning next node

for curr.next != nil && curr.next.value != val { curr = curr.next }

move curr to node before the one to delete

if curr.next != nil { curr.next = curr.next.next }

skip the node with target value to delete it

Output

Original list: 1 -> 2 -> 3 -> 4 -> null

After deleting 3: 1 -> 2 -> 4 -> null

Complexity Analysis

Time: O(n) because we may need to check each node once to find the value

Space: O(1) because we only use a few pointers and no extra data structures

vs Alternative: Compared to rebuilding the list without the value, this approach changes links directly, saving time and space

Edge Cases

empty list

function returns nil without error

DSA Go

if head == nil { return head }

value is at head node

head is removed by returning head.next

DSA Go

if head.value == val { return head.next }

value not found in list

list remains unchanged

DSA Go

if curr.next != nil { curr.next = curr.next.next }

When to Use This Pattern

When asked to remove a node by value in a linked list, look for the pattern of finding the node before it and changing pointers to skip it.

Common Mistakes

Mistake: Trying to delete the node without checking if it's the head node
Fix: Add a check to handle deletion when the head node contains the target value

Mistake: Not stopping traversal when the node before the target is found
Fix: Use a loop condition that stops at the node before the one to delete

Summary

Deletes the first node with a given value from a linked list.

Use when you need to remove a specific value from a singly linked list.

The key is to find the node before the target and change its next pointer to skip the target node.

Practice

1. What is the main purpose of the Delete Node by Value operation in a linked list?

easy

A. To find the position of a node with the given value

B. To remove the first node that contains the given value from the list

C. To add a new node with the given value at the end of the list

D. To reverse the linked list

Solution

1. Step 1: Understand the operation's goal

   The operation aims to remove a node that matches a specific value from the linked list.

2. Step 2: Identify the correct description

   Removing the first node with the given value matches To remove the first node that contains the given value from the list, while others describe different operations.

3. Final Answer:

   To remove the first node that contains the given value from the list -> Option B

4. Quick Check:

   Delete node by value means remove matching node [OK]

Hint: Delete node by value means remove matching node [OK]

Common Mistakes:

• Confusing deletion with insertion
• Thinking it finds position instead of deleting
• Assuming it reverses the list

2. Which of the following Go code snippets correctly deletes the first node with value 5 from a singly linked list starting at head?

easy

A. if head == nil { return head } if head.val == 5 { return head.next } curr := head for curr.next != nil && curr.next.val != 5 { curr = curr.next } if curr.next != nil { curr.next = curr.next.next } return head

B. if head == nil { return nil } curr := head for curr != nil && curr.val != 5 { curr = curr.next } if curr != nil { curr = curr.next } return head

C. if head.val == 5 { head = nil } curr := head for curr.next != nil { if curr.next.val == 5 { curr.next = nil } curr = curr.next } return head

D. curr := head for curr.next != nil { if curr.val == 5 { curr = curr.next.next } curr = curr.next } return head

Solution

1. Step 1: Check for empty list and head node value

   if head == nil { return head } if head.val == 5 { return head.next } curr := head for curr.next != nil && curr.next.val != 5 { curr = curr.next } if curr.next != nil { curr.next = curr.next.next } return head correctly returns head.next if head's value is 5, handling deletion at the start.

2. Step 2: Traverse and delete node with value 5

   if head == nil { return head } if head.val == 5 { return head.next } curr := head for curr.next != nil && curr.next.val != 5 { curr = curr.next } if curr.next != nil { curr.next = curr.next.next } return head uses a loop to find the node before the target and updates links properly; others miss this or incorrectly update pointers.

3. Final Answer:

   Option A code correctly deletes the first node with value 5 -> Option A

4. Quick Check:

   Proper pointer update and head check = if head == nil { return head } if head.val == 5 { return head.next } curr := head for curr.next != nil && curr.next.val != 5 { curr = curr.next } if curr.next != nil { curr.next = curr.next.next } return head [OK]

Hint: Check head value first, then update next pointers carefully [OK]

Common Mistakes:

• Not handling deletion when head node matches
• Incorrect pointer updates causing list break
• Skipping traversal or deleting wrong node

3. Given the linked list: 1 -> 3 -> 5 -> 7 -> 5 -> null, what will be the list after deleting the first node with value 5 using the standard delete-by-value function?

medium

A. 1 -> 3 -> 5 -> 7 -> null

B. 1 -> 3 -> 7 -> null

C. 3 -> 5 -> 7 -> 5 -> null

D. 1 -> 3 -> 7 -> 5 -> null

Solution

1. Step 1: Identify the first node with value 5

   The first 5 appears as the third node in the list (1 -> 3 -> 5).

2. Step 2: Remove that node and link previous to next

   Deleting this node links 3 directly to 7, resulting in 1 -> 3 -> 7 -> 5 -> null.

3. Final Answer:

   1 -> 3 -> 7 -> 5 -> null -> Option D

4. Quick Check:

   First 5 removed, rest unchanged = 1 -> 3 -> 7 -> 5 -> null [OK]

Hint: Delete only first matching node, keep rest intact [OK]

Common Mistakes:

• Deleting all nodes with value 5
• Removing wrong node (second 5)
• Breaking list links incorrectly

4. The following Go code attempts to delete a node by value from a singly linked list but causes a runtime panic. What is the error?

func deleteNode(head *Node, val int) *Node {
    curr := head
    for curr.next != nil {
        if curr.next.val == val {
            curr.next = curr.next.next
            break
        }
        curr = curr.next
    }
    return head
}

medium

A. It does not handle deletion if the head node contains the value

B. It causes infinite loop due to missing curr update

C. It tries to access curr.next.val without checking if curr.next is nil

D. It deletes all nodes with the value instead of just the first

Solution

1. Step 1: Analyze head node deletion case

   The code starts from head and checks curr.next, so if head itself has the value, it is not deleted.

2. Step 2: Confirm no panic cause

   The loop condition curr.next != nil prevents nil pointer dereference, so no panic from accessing curr.next.val.

3. Final Answer:

   It does not handle deletion if the head node contains the value -> Option A

4. Quick Check:

   Head node deletion missing = It does not handle deletion if the head node contains the value [OK]

Hint: Always check head node separately before looping [OK]

Common Mistakes:

• Ignoring head node deletion
• Assuming panic from nil pointer without checking
• Not breaking after deletion causing multiple deletes

5. You have a singly linked list with nodes containing values: 2 -> 4 -> 6 -> 4 -> 8 -> null. You want to delete all nodes with value 4 using a single pass delete-by-value function that deletes only the first matching node per call. How many times must you call this function to remove all 4s, and what will the list look like after all deletions?

hard

A. Call twice; final list: 2 -> 4 -> 6 -> 8 -> null

B. Call once; final list: 2 -> 6 -> 4 -> 8 -> null

C. Call twice; final list: 2 -> 6 -> 8 -> null

D. Call once; final list: 2 -> 6 -> 8 -> null

Solution

1. Step 1: Understand single pass deletion removes only first matching node

   Each call deletes only the first node with value 4 found.

2. Step 2: Count calls needed and final list

   First call removes the first 4 (second node), list becomes 2 -> 6 -> 4 -> 8 -> null. Second call removes the next 4 (third node), list becomes 2 -> 6 -> 8 -> null.

3. Final Answer:

   Call twice; final list: 2 -> 6 -> 8 -> null -> Option C

4. Quick Check:

   Two calls remove both 4s = Call twice; final list: 2 -> 6 -> 8 -> null [OK]

Hint: One call deletes one node; repeat until all removed [OK]

Common Mistakes:

• Assuming one call removes all matching nodes
• Miscounting nodes deleted per call
• Confusing node positions after deletions