What is the output of following function for start pointing to first node of following linked list?1->2->3->4->5->6void fun(struct node* start){  if(start == NULL)    return;  printf("%d  ", start->data);   if(start->next != NULL )    fun(start->next->next);  printf("%d  ", start->data);}1 4 6 6 4 11 3 5 5 3 11 3 5 1 3 51 2 3 5

What will be the output of the following code snippet for the list 1->2->3->4->5->6?void solve(struct node* start){if(start == NULL)return;printf("%d ", start->data);if(start->next != NULL )solve(start->next->next);printf("%d ", start->data);}*2 points1 2 3 4 5 61 3 5 5 3 61 3 5 1 3 52 4 6 1 3 5

The following C function takes a single-linked list of integers as a parameter and rearranges the elements of the list. The function is called with the list containing the integers 1, 2, 3, 4, 5, 6, 7 in the given order. What will be the contents of the list after the function completes execution?struct node{  int value;  struct node *next;};void rearrange(struct node *list){  struct node *p, * q;  int temp;  if ((!list) || !list->next)      return;  p = list;  q = list->next;  while(q)  {     temp = p->value;     p->value = q->value;     q->value = temp;     p = q->next;     q = p?p->next:0;  }}1,2,3,4,5,6,72,1,4,3,6,5,72,3,4,5,6,7,11,3,2,5,4,7,6

How is the last node identified in a doubly linked list?It points to `None`.It has a special flag.It has a reference to the first node.It has a pointer to the previous node.

In a doubly linked list, what’s the “tail” of a linked list?It’s the node with the pointer to the previous node equal to NULLIt’s the node with the pointer to the next node equal to NULL

void myFun(Node* head) {Node* current = head;while (current != nullptr) {if((current->data)%2==0)std::cout << current->data << " "; current = current->next;}std::cout << std::endl; }Output of above code in linked list