Question

What is the output of following functoin for start pointing to first node of following linked list? 1 ->2 -> 3 -> 4 -> 5 -> 6 ->NULL void fun(struct node* start) { if(start == NULL) return; printf("%d ", start->data); if(start->next != NULL ) fun(start->next->next); printf("%d ", start->data); }

Answer 1

The output of the given function for the provided linked list is '1 3 5 5 3 1'. The function prints certain nodes' data twice due to its recursive and printing statements.

Step-by-step explanation:

The given function is part of a C program dealing with a singly linked list. The function named fun takes a pointer to the first node of a linked list as an argument. The function's output is generated by a combination of recursive calls and print statements that result in certain values from the linked list being printed out.

The function first checks if the given start pointer is NULL, indicating the end of the linked list. If it is not NULL, the function prints the data of the current node. Then it checks if there is a next node. If there is, the function makes a recursive call with the "next of next" node as the argument. After the recursive call(s), the data of the current node is printed again.

For the provided linked list (1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL), the output of the function when called with the start pointing to the first node would be:

1 3 5 5 3 1

This is because the function will print the first node, skip the second node, print the third, skip the fourth, and so on. Due to the recursive structure, after reaching the end of the list, the function will backtrack, printing the data of the nodes in reverse order on its way back.