Linked list is a linear data structure used to store elements of similar data type.

Each node in a list consists of at least two parts:

- data value
- pointer to the next node

Linked list can be formed in two ways:

- with a root element
- without a root

Checking if the given Linked List is empty depends on the ways Linked List has been formed. We have covered both approaches.

# With a root

The root acts as an element which is always present even if the list is empty.

The other use of having a root in list is that we can link the last element back to the root forming a cycle. According to this concept, **next** of any node is never NULL.

Pseudocode:

```
bool isEmpty(node *root){
if (root->next == root) return true;
else return false;
}
```

The implementation is given below:

```
#include<bits/stdc++.h>
using namespace std;
class Node{
public:
int data;
Node *next;
Node(){
data=0;
next=NULL;
}
};
class linked_list{
Node *root;
public:
linked_list(){
root=NULL;
}
Node* getRoot(){
return root;
}
void add_node(int n){
Node *temp = new Node();
temp->data = n;
temp->next = NULL;
if(root == NULL){
root=temp;
root->next=root;
}
else{
Node *last=root;
while(last->next!=root){
last=last->next;
}
temp->next=root;
last->next=temp;
}
}
void printList(){
Node *temp=root;
if(temp!=NULL){
do{
cout<<temp->data<<" ";
temp = temp->next;
} while(temp!=root);
}
}
bool isEmpty(){
if(root->next==root && root->data==0) return true;
return false;
}
};
int main(){
linked_list l1;
l1.add_node(5);
l1.add_node(10);
l1.add_node(15);
if(l1.isEmpty()) cout<<"The list is empty!\n";
else {
cout<<"The list is not empty! List contains:\n";
l1.printList();
}
return 0;
}
```

**Output:**

# without root

Without a root, the list pointer is NULL when the list is empty.

Pseudocode:

```
bool isEmpty(node *list){
if (list == NULL) return true;
else return false;
}
```

Implementation of Linked List without root is given below:

```
#include<bits/stdc++.h>
using namespace std;
class Node
{
public:
int data;
Node *next;
} ;
void push(Node** head_ref, int new_data){
Node* new_node = new Node();
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
bool isEmpty( Node** list){
if((*list)==NULL) return true;
return false;
}
void printList(Node *node)
{
while (node != NULL)
{
cout<<node->data<<" ";
node = node->next;
}
}
int main(){
Node *list=NULL;
if(isEmpty(&list)) cout<<"List is Empty"<<endl;
else{
cout<<"List is not empty! List contains:"<<" ";
printList(list);
}
// Inserting some elements in list
push(&list,8);
push(&list,4);
if(isEmpty(&list)) cout<<"List is Empty"<<endl;
else{
cout<<"List is not empty! List contains:"<<" ";
printList(list);
}
return 0;
}
```

**Output:**

**Complexity:**

The complexity to check whether the list is empty or not in both the approaches is same i.e O(1)