Expect the output for the following code of 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);}Options2 3 1 4 5 61 2 3 4 5 61 4 2 3 5 61 3 5 5 3 1
Question
Expect the output for the following code of 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);}Options2 3 1 4 5 61 2 3 4 5 61 4 2 3 5 61 3 5 5 3 1
Solution
The output for the given code will be 1 3 5 5 3 1.
Here's the step by step explanation:
-
The function
solveis called with the head of the list (node with value 1). -
It checks if the start is NULL, which it isn't, so it prints the data of the start node, which is 1.
-
It then checks if the next node is not NULL, which it isn't, so it calls the
solvefunction with the node two steps ahead (node with value 3). -
The same process repeats for the nodes with values 3 and 5.
-
When it tries to call the
solvefunction with the node two steps ahead of the node with value 5, it finds that it is NULL, so it doesn't call the function and instead goes back to the previous call of the function. -
It then prints the data of the current node again, which is 5.
-
It goes back to the previous calls of the function and prints the data of the nodes with values 3 and 1 again.
So, the output is 1 3 5 5 3 1.
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