# Print the Elements of a Linked List

### Problem Statement :

```This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print.

Function Description:

Complete the printLinkedList function in the editor below.

Print:
1. For each node, print its  value on a new line (console.log in Javascript).

Input Format:

The first line of input contains n, the number of elements in the linked list.
The next n lines contain one element each, the data values for each node.

Note: Do not read any input from stdin/console. Complete the printLinkedList function in the editor below.

Constraints:
1. 1<=n<=1000
2. 1<=list[i]<=1000```

### Solution :

```                            ```Solution in C :

In C:
//In hacker rank as the solution the program is too big
//all u need is to code this function.

while(ptr!=NULL)
{
printf("%d\n",ptr->data);
ptr=ptr->next;
}

}

In C++:
//The rest of the code is already coded.
//As the code is too big all you need is complete the following part.

{
}
}

In Java:
//The following part is all you to complete, the rest is

}

In Python 3:

# The following part of code is all you need to
# complete the challenge in hacker rank

while (temp.next != None):
print(temp.data)
temp = temp.next
print(temp.data)```
```

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