261 ICS Fall 2002 Data Structures Midterm 1 Answers
Comments on the grading are given in smaller type after the answers.
Part I Version A: 1. C 2. C 3. B 4. B 5. B 6. E 7. C 8. C 9. B Version B: 1. B 2. C 3. B 4. D 5. D 6. E 7. C 8. A 9. B Part II. 10. (Here "3x" means 33 on Version A and 37 on Version B.) We add 3x as a right child of the node containing 30. Then the deepest imbalanced node is 20, with a balance of +2, and its right child has balance -1. Therefore we do a double left rotation at that node, obtaining 60 30 20 10 3x 40 50 70 80
(On many of your answers you will see lines drawn enclosing nodes that are about to be involved in a rotation; I drew these to help me understand your answers.) Here are examples of credit for answers that were not completely correct. 10: The insertion was done correctly, and work was shown, but there was an error in an intermediate step. 8: The insertion was done correctly except that the 50 was omitted in the final answer, and work was shown. 6: The double rotation was done one level too high in the tree. This yields the following tree, which does turn out to be a valid AVL tree. 40 20 10 30 3x 50 70 60 80
7: As above, but the deepest imbalanced node was correctly identified as 20. 6: The insertion was done correctly except that 70 was omitted from the tree, and therefore another rotation had to be done to make the tree an AVL tree. 5: The insertion was done incorrectly by splicing in the 3x above the 30, but then the correct rotation was done given this starting point, and the final tree was an AVL tree. 4: Doing the insertion correctly and then doing an incorrect series of rotations that I could not follow, or doing a pseudosplay at 3x, but still winding up with an AVL tree. 2: Doing the binary search tree insertion correctly, but then thinking that the tree was balanced and not doing any rotations.
ICS 261 Fall 2002
Midterm 1 Answers
1: Doing a sequence of rotations that I could not follow, and winding up with a tree that was not an AVL tree or binary search tree.
11. a) i. No ii. Yes iii. Yes, T4 could have a red root (since its original parent C was black), in which case in the new tree the root of T4 and its parent B would be two consecutive red nodes on a path from a leaf to the root. b) i. Yes ii. No be red. iii. Yes, A is a red node with a red parent. Also, B's parent could
c) i. Yes ii. Yes iii. Yes, if B's parent was also red, B and its parent would be two consecutive red nodes on a path from a leaf to the root.
For parts (i) and (ii) of each case, a simple yes or no sufficed. Each of these was worth 1 point, and I tended to give this point for the correct yes-no answer even if there was an explanation that didn't make much sense to me. For part (iii) of each case, required I that the specific problem be identified. This part was worth 2 points for each case. Examples of credit are: 0: Saying there is a problem, but giving an incorrect problem or just repeating a problem identified in part (i) or (ii). 1: Saying there is a problem and giving a partially correct explanation. Examples are saying "T1 , T2 , T3 , T4 could have red roots" for case (a), or saying "A should be black" in case (b). 2: Repeating one of the problems identified in part (i) or (ii), but also giving a new problem.
12. The key observation is that a tree on 2k keys will be an augmented perfect tree if and only if one of its subtrees is a perfect tree on 2k-1 - 1 nodes and the other subtree is an augmented perfect tree on 2k-1 nodes. This happens if and only if a) the first key inserted is 2k-1 , the left subtree is a perfect tree on 2k-1 - 1 nodes, and the right subtree is an augmented perfect tree on 2k-1 nodes, or b) the first key inserted is 2k-1 + 1, the left subtree is an augmented perfect tree on 2k-1 nodes, and the right subtree is a perfect tree on 2k-1 - 1 nodes. The probability that any particular key is the first to be inserted is 2-k . For each case above, if the given key is the first inserted, the left and right subtrees are independent so the probability that they both have the required shapes is qk-1 rk-1 . Thus the probability of each case is 2-k qk-1 rk-1 Adding the probability of the two cases gives qk = 2 2-k qk-1 rk-1 = qk-1 rk-1 . 2k-1
The final formula was worth 9 points, and the explanation was worth 3 points. Most of you will see a notation of the form a, b below your answer, where...