CMPE231 DATA STRUCTURES FINAL EXAMINATION / FALL 2010

Transcription

1 EASTERN MEDITERRANEAN UNIVERSITY Computer Engineering Department CMPE231 DATA STRUCTURES FINAL EXAMINATION / FALL 2010 Lecturers: Prof.Dr.Marifi Güler (group 1) Prof.Dr.Erden Başar (group 2) January 14, 2010 Duration: 120 minutes Name, Surname KEY Student ID # Group # 1 2 Please, check your exam sheet and make sure that it contains 8 questions. In case of any missing pages, inform the invigilator IMMEDIATELY! Grading: Q1. / 14 Q2. / 16 Q3. / 12 Q4. / 6 Q5. / 8 Q6. / 12 Q7. / 16 Q8. / 16 Total If an Answer Box is provided in a question, you MUST give your answer (and nothing else) in the corresponding box. Otherwise, you will get NO MARKS from that question!!!

2 1)(14 points) Consider the following binary search tree: K G N B L T E P D S a.) Give the preorder traverse of the tree KGBEDNLTPS b.) Give the inorder traverse of the tree BDEGKLNPST c.) Draw the binary search tree when the node with the value N is deleted. K G P B L T E S D

3 2) (16 points) A) Draw the binary tree representing the following arithmetic expression: (D + F) $ A $ (B - C) / E * G where $ has the highest precedence and right to left associativity. where $ has the highest precedence and right to left associativity. * / G $ E + $ D F A - B)What is the postorder traversal of the tree? B C DF+ABC-$$E/G* C) Using below algorithm(eval), trace and find the resulted value of the postfix expression that you obtained after the postorder traversal of the tree at step B. (Data values are A = 1, B = 4, C = 2, D = 6, E = 3, F = 3, G = 5) stack=the empty stack while (not end of input) symb = next input character if (symb is an operand) then push(stack, symb) else / * symb is an operator */ opnd2 = pop(stack)

4 opnd1 = pop(stack) value = result of appliying symb to opnd1 and opnd2 push(stack, value) /* end while */ return (pop(stack) D F + A B C - $ $ E / G * $ $ 3 / 5 * symb stack opnd1 opnd2 value $ 1 $ 2 1 $ $ / 9 / * 3 * 5 15

5 3)(12 points)assume that the linked list pointed to by list contains the data as shown in the figure. list What will the contents of the list be after calling the function modify( ) (see definition below) as modify(list);? void fun(struct node *, int ); struct node int info; struct node * next; ; void modify(struct node * list) struct node * qq; int x = 4; for(qq = list; qq!= NULL; qq = qq -> next) if(qq -> info = = 2) fun(qq, ++x); void fun(struct node * p, int x) struct node * q; q = (struct node *) malloc(sizeof(struct node)); q -> info = x; q -> next = p -> next; p -> next = q;

6 4)(6 points) Give(Draw) the binary search tree corresponding to a set of items presented as follows:

7 5)(8 Points) Find (draw) the binary tree whose Postorder Traversal is A B C - D E F * $ * + And whose Inorder Traversal is A + B C * D $ E * F + A * - $ B C D * E F 6)(12 Points) By using a circular linked list, a queue may be specified by a single pointer pque to that list. Node(pque) is the rear(last Node) of the queue and the following node is its front(first Node). pque First Node Last Node The following program does the necessary deletion from circular queue. However, some parts of the code are missing (underlined as Fill in the incomplete parts(3 lines). struct node int info; struct node *next; ; struct node *NODEPTR; int delque(nodeptr *pque) int x; NODEPTR q; If (empty(pque) == TRUE) printf( Queue underflow \ ); exit(1);

8 q = (*pque) next; x=q->info; if (q == *pque ) *pque = NULL; else *pque) next = q next; freenode(q); return (x); 7)(16 Points)Draw the tree structure and display the output of the given program. struct allinfo Draw tree char data; int c; ; struct nodetype struct nodetype *left ; struct allinfo info; struct nodetype *right; G 1 / ; typedef struct nodetype NODE; typedef NODE *NODEPTR; / K 1 / void main() NODEPTR root, p, q; root=null; char one, alf[9]='d', 'B', 'C', 'D', 'M', 'A', 'B', 'G', 'K'; int i=0; one = alf[i]; root = maketree(one); do one = alf[++i]; p=q=root; while((one!= p->info.data) && q!=null) p=q; D 2 / M 1 B 2 Output of the program X Traversing Data = M C = 1 Data = K C = 1 Data = G C = 1 Data = D C = 2 Data = C C = 1 Data = B C = 2 Data = A C = 1 / C 1 / / A 1 /

9 if (one > p->info.data) q = p->left; else q = p->right; if (one == p->info.data) p->info.c = p->info.c+1; else q=maketree(one); if (one > p->info.data)p->left = q; else p->right = q; while(i!=8); printf("x Traversing \n "); xtrav(root); void xtrav(nodeptr tree) if (tree!= NULL) xtrav(tree->left); printf("data=%c C=%d\n", tree->info.data, tree->info.c); xtrav(tree->right); NODEPTR maketree(char x) NODEPTR p; p=(nodeptr) malloc(sizeof(struct nodetype)); p->info.data = x; p->info.c = 1; p->left = NULL; p->right = NULL; return(p); 8)(16 Points)Complete the following main program to create a Doubly circular sorted linked list structure which will have one alphabet character in each node as shown below. At the end list will point the address of last character Z. struct node struct node *left; char letter; struct node *right; ; typedef struct node *NODEPTR; list A B C D Z Assume that getnode() function is already written with the following prototype. NODEPTR getnode(void); void main() NODEPTR p, list, save; /* character A initially is inserted into doubly circular linked list as follows */ p=getnode();

10 p->letter= A ; list=p; p->right=p; p->left=p; save=list; for(int i = 1 ; i < 26 ; i++) p = getnode(); p letter= A + i; p left = save; save right = p; save = p; list left = save; save right = list ; list = p; NODEPTR getnode() NODEPTR q; q = (NODEPTR) malloc(sizeof(struct node)); return(q); Hint : character B can be obtained as B = A + 1