Question

(ii) [6 marks] Assume that we have an empty stack S and an empty queue Q. Given a series of stack operations on S as below: Push(S,10), Push(S, 7), Push(S, 23), Pop(S), Push(S, 9), Pop(S), Pop(S) Output the element returned by each Pop operation. Given a series of queue operations on Q as below: Enqueue(0,10),Enqueue(Q,20), Enqueue(0,33), Dequeue(Q), Enqueue(Q,55), Dequeue(Q), Dequeue(Q) Output the element returned by each Dequeue operation. (iii) [8 marks] Given an empty binary search tree T, draw the binary search tree T after inserting 25, 20, 50, 15, 60, 30, 18 into T.

Answer 1

$\blacklozenge$ We know that in a Stack when we insert or push elements it follows last-in-first-out ( LIFO ) rule, that is, last inserted element will be the first to be deleted or popped. This is because the Stack pointer, top of stack ( TOS ), always points to the new element being inserted.

So, after push(S,10), push(S,7) and push(S,23) where S is the stack and the other parameter is the element to be inserted, our Stack S would look like

23

7

10

And TOS will point to 23.

then first pop(S) will give 23. And stack will become, for us,

7

10

And TOS points to 7. Then push(S,9) is performed. Stack becomes

9

7

10

And TOS points to 9.

Then Second pop(S) will give 9. TOS will then point to 7 again.

And third pop(S) will give 7. The stack at the end will become

10

$\blacklozenge$ We know that in a queue when we insert or enqueue elements it follows first-in-first-out ( FIFO ) rule, that is, first inserted element will be the first to be deleted or dequeued. This is because the queue maintains two pointers,

front pointer (f) : points to the first element to be deleted and,

rear pointer(r) : points where the next element will be enqueued,

So, after enqueue(Q,10) , enqueue(Q,20) and enqueue(Q,33) our queue Q will look like :

10

20

33

f points to 10, and r points to 33.

first dequeue(Q) will give 10. And queue will be

20

33

where f will now point to 20 and r will point to 33.

After enqueue(Q,55) queue will become

20

33

55

where f will point to 20 and r will now point to 55.

Then second dequeue(Q) will give 20. And queue will be

33

55

where f will now point to 33 and r will point to 55.

Then third dequeue(Q) will give 33. And queue will be

55

where f will also point to 55 along with r .

$\blacklozenge$ BInary Search Tree is based on the fact that the left child of a node is always less than the node and the right child is always greater than the node, that is, the left subtree of the root will consist of nodes lesser than the root and right subtree will consist of nodes greater than the root.

25 20 50 15 30 60 18

Here, we can see that first element we inserted was 25, which becomes our root node as the tree was initially empty.

Second we insert 20, which goes to the left of root as it's less than 25.

Then we add 50 which is greater than root 25 and thus, it goes to the right of root node.

Then we add 15, which being less than 25 goes to left subtree, and then we check if 15 is less than or greater than 20, as it's less than 20 it becomes left child of 20.

Then we are adding 60 which is greater than root so, it goes to the right subtree. Then we compare it with 50 and as 60 is greater it becomes the right child of 50.

Then we add 30, which is greater than 25, it goes to right subtree, and then we compare it with 50 30 being less than 50, becomes the left child of 50.

At last we add 18 which being less than 25 goes to the left subtree and then we compare it with 20, which is greater than 18 so 18 should have been in the left of 20, but as 15 is already there so it is again compared with 15 and 18 is greater than 15 so, 18 becomes the right child of 15.

And thus. BST is completed.