C++ Program for Deleting a Node in a Linked List
Last Updated :
26 Aug, 2024
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Write a C++ program to delete a node from the given link list.
Examples
Input: Linked List: 10 -> 20 -> 30 -> 40 -> 50, Position to delete: 3
Output: 10 -> 20 -> 40 -> 50
Explanation: The node at position 3 is removed. The list then connects node 20 directly to node 40.Input: Linked List: 5 -> 15 -> 25, Position to delete: 1
Output: 15 -> 25
Explanation: The head node is deleted. The head pointer is updated to point to the next node.
Types of Delete Operation on a Linked List
There are three basic types of delete operations on a singly linked list:
1. Delete the Head Node
When deleting the head node, we need to adjust the head pointer to point to the next node otherwise, the whole linked list can be lost.
Following is the approach to remove the head node of the linked list:
- Store the current head node in a temporary variable.
- Move the head pointer to the next node.
- Delete the temporary node.
Implementation
// C++ program to show how to delete a head node from
// the linked list
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int data) {
this->data = data;
this->next = nullptr;
}
};
// Function to delete the head node
void deleteHead(Node *&head) {
if (head == nullptr) {
cout << "List is empty, nothing to delete." << endl;
return;
}
// Store current head in a temporary variable
Node *temp = head;
// Move head to the next node
head = head->next;
// Delete the old head node
delete temp;
}
int main() {
// Create a simple linked list: 3 -> 12 -> 15 -> 18
Node *head = new Node(3);
head->next = new Node(12);
head->next->next = new Node(15);
head->next->next->next = new Node(18);
cout << "Original List: ";
Node *temp = head;
while (temp != nullptr) {
cout << temp->data << " -> ";
temp = temp->next;
}
cout << "NULL" << endl;
deleteHead(head); // Deleting the head node
cout << "List after deleting head: ";
temp = head;
while (temp != nullptr) {
cout << temp->data << " -> ";
temp = temp->next;
}
cout << "NULL" << endl;
return 0;
}
// C++ program to show how to delete a head node from// the linked listusing namespace std;class Node {public: int data; Node *next; Node(int data) { this->data = data; this->next = nullptr; }};// Function to delete the head nodevoid deleteHead(Node *&head) { if (head == nullptr) { cout << "List is empty, nothing to delete." << endl; return; } // Store current head in a temporary variable Node *temp = head; // Move head to the next node head = head->next; // Delete the old head node delete temp;}int main() { // Create a simple linked list: 3 -> 12 -> 15 -> 18 Node *head = new Node(3); head->next = new Node(12); head->next->next = new Node(15); head->next->next->next = new Node(18); cout << "Original List: "; Node *temp = head; while (temp != nullptr) { cout << temp->data << " -> "; temp = temp->next; } cout << "NULL" << endl; deleteHead(head); // Deleting the head node cout << "List after deleting head: "; temp = head; while (temp != nullptr) { cout << temp->data << " -> "; temp = temp->next; } cout << "NULL" << endl; return 0;}Output
Original List: 3 -> 12 -> 15 -> 18 -> NULL List after deleting head: 12 -> 15 -> 18 -> NULL
Time Complexity: O(1)
Auxiliary Space: O(1)
2. Delete the Tail Node
To delete the tail node, traverse the list to find the second-to-last node, and then set its next pointer to NULL.
Following is the approach to remove the head node of the linked list:
- Traverse the list to find the second-to-last node.
- Set the next pointer of the second-to-last node to NULL.
- Delete the last node.
Implementation
// C++ program to show how to delete a head node from
// the linked list
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int data) {
this->data = data;
this->next = nullptr;
}
};
void deleteTail(Node*& head) {
// Check if the list is empty
if (head == nullptr) {
cout << "List is empty, nothing to delete." << endl;
return;
}
// Check if it contains only one element
if (head->next == nullptr) {
delete head;
head = nullptr;
return;
}
// Traverse to the tail node
Node* temp = head;
while (temp->next->next != nullptr) {
temp = temp->next;
}
// delete the tail node
delete temp->next;
temp->next = nullptr;
}
int main() {
// Create a simple linked list: 3 -> 12 -> 15 -> 18
Node *head = new Node(3);
head->next = new Node(12);
head->next->next = new Node(15);
head->next->next->next = new Node(18);
cout << "Original List: ";
Node *temp = head;
while (temp != nullptr) {
cout << temp->data << " -> ";
temp = temp->next;
}
cout << "NULL" << endl;
deleteTail(head); // Deleting the tail node
cout << "List after deleting head: ";
temp = head;
while (temp != nullptr) {
cout << temp->data << " -> ";
temp = temp->next;
}
cout << "NULL" << endl;
return 0;
}
// C++ program to show how to delete a head node from// the linked listusing namespace std;class Node {public: int data; Node *next; Node(int data) { this->data = data; this->next = nullptr; }};void deleteTail(Node*& head) { // Check if the list is empty if (head == nullptr) { cout << "List is empty, nothing to delete." << endl; return; } // Check if it contains only one element if (head->next == nullptr) { delete head; head = nullptr; return; } // Traverse to the tail node Node* temp = head; while (temp->next->next != nullptr) { temp = temp->next; } // delete the tail node delete temp->next; temp->next = nullptr;}int main() { // Create a simple linked list: 3 -> 12 -> 15 -> 18 Node *head = new Node(3); head->next = new Node(12); head->next->next = new Node(15); head->next->next->next = new Node(18); cout << "Original List: "; Node *temp = head; while (temp != nullptr) { cout << temp->data << " -> "; temp = temp->next; } cout << "NULL" << endl; deleteTail(head); // Deleting the tail node cout << "List after deleting head: "; temp = head; while (temp != nullptr) { cout << temp->data << " -> "; temp = temp->next; } cout << "NULL" << endl; return 0;}Output
Original List: 3 -> 12 -> 15 -> 18 -> NULL List after deleting head: 3 -> 12 -> 15 -> NULL
Time Complexity: O(N), where N is the number of nodes in the list.
Auxiliary Space: O (1)
3. Delete a Node at a Specific Position
To delete a node at a specific position, traverse the list to reach the node just before the target position, adjust pointers to skip the target node, and delete it.
Following is the approach to remove the head node of the linked list:
- If the position is 1, delete the head. Otherwise,
- Traverse the list to reach the node before the target position.
- Adjust pointers to bypass the target node and point to then next of the target node.
- Delete the target node.
Implementation
// C++ program to show how to delete a head node from
// the linked list
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int data) {
this->data = data;
this->next = nullptr;
}
};
// Function to delete a node at a specific position
void deleteAtPosition(Node*& head, int pos) {
if (head == nullptr || pos < 1) {
cout << "Invalid position or list is empty." << endl;
return;
}
// Deleting head node
if (pos == 1) {
Node* temp = head;
head = head->next;
delete temp;
return;
}
// Traverse to the node before the target position
Node* current = head;
for (int i = 1; i < pos - 1 && current != nullptr; i++) {
current = current->next;
}
// If position is greater than the number of nodes
if (current == nullptr || current->next == nullptr) {
cout << "Position exceeds list length." << endl;
return;
}
// Delete the node at the target position
Node* temp = current->next;
current->next = temp->next;
delete temp;
}
int main() {
// Create a simple linked list: 3 -> 12 -> 15 -> 18
Node *head = new Node(3);
head->next = new Node(12);
head->next->next = new Node(15);
head->next->next->next = new Node(18);
cout << "Original List: ";
Node *temp = head;
while (temp != nullptr) {
cout << temp->data << " -> ";
temp = temp->next;
}
cout << "NULL" << endl;
deleteAtPosition(head, 2); // Deleting node at position 3
cout << "List after deleting head: ";
temp = head;
while (temp != nullptr) {
cout << temp->data << " -> ";
temp = temp->next;
}
cout << "NULL" << endl;
return 0;
}
// C++ program to show how to delete a head node from// the linked listusing namespace std;class Node {public: int data; Node *next; Node(int data) { this->data = data; this->next = nullptr; }};// Function to delete a node at a specific positionvoid deleteAtPosition(Node*& head, int pos) { if (head == nullptr || pos < 1) { cout << "Invalid position or list is empty." << endl; return; } // Deleting head node if (pos == 1) { Node* temp = head; head = head->next; delete temp; return; } // Traverse to the node before the target position Node* current = head; for (int i = 1; i < pos - 1 && current != nullptr; i++) { current = current->next; } // If position is greater than the number of nodes if (current == nullptr || current->next == nullptr) { cout << "Position exceeds list length." << endl; return; } // Delete the node at the target position Node* temp = current->next; current->next = temp->next; delete temp;}int main() { // Create a simple linked list: 3 -> 12 -> 15 -> 18 Node *head = new Node(3); head->next = new Node(12); head->next->next = new Node(15); head->next->next->next = new Node(18); cout << "Original List: "; Node *temp = head; while (temp != nullptr) { cout << temp->data << " -> "; temp = temp->next; } cout << "NULL" << endl; deleteAtPosition(head, 2); // Deleting node at position 3 cout << "List after deleting head: "; temp = head; while (temp != nullptr) { cout << temp->data << " -> "; temp = temp->next; } cout << "NULL" << endl; return 0;}Output
Original List: 3 -> 12 -> 15 -> 18 -> NULL List after deleting head: 3 -> 15 -> 18 -> NULL
Time Complexity: O(N), where N is the number of nodes in the list.
Auxiliary Space: O (1)
Conclusion
Deleting a node from a linked list is a common operation that requires careful handling of pointers to avoid breaking the list structure. Depending on the position of the node to be deleted (head, tail, or a specific position), the approach varies slightly. These examples demonstrate how to handle different deletion scenarios in a singly linked list effectively.
