1. Foundations of Data Structures Practical Session #7 AVL Trees 2

2. AVL Tree properties Height-Balance Property AVL Tree AVL Interface AVL Height 2

3. AVL Tree example 3 14 17 7 4 53 11 12 813

4. Question 1 Insert the following sequence of integers into an empty AVL tree: 14, 17, 11, 7, 53, 4, 13 4 14 1711 753 4

5. right A single right rotation of ’11’ is executed to rebalance the tree: Insert 13 5 14 177 45311 13

6. Now insert 12 6 14 177 45311 13 12 The sub-tree of 11 is unbalanced. Double rotation: right and then left.

7. After right rotation of ’13’ Now left rotate ’11’ 7 14 177 45311 12 13

8. After left rotation of ’11’ Now balanced! 8 14 177 45312 1311

9. Now insert 8 9 14 177 45312 1311 8 The sub-tree of 7 is unbalanced. Required double rotation: right and then left.

10. After right rotation of ’12’ Now left rotate ‘7’ 10 14 177 45311 128 13

11. Now balanced! 11 14 17 7 4 53 11 12 813

12. Remove 53 12 14 17 7 4 53 11 12 813

13. Unbalanced! Right rotate ’14’ 13 14 17 7 4 11 12 813

14. Balanced! Remove 11 14 17 7 4 11 128 13

15. Remove 11 Replace it with the maximum in its left branch 15 14 17 7 4 11 128 13

16. Remove 8 16 14 17 7 4 8 12 13

17. Unbalanced! Required double rotatation 17 14 17 4 7 12 13

18. After right rotation of ‘14’ 18 14 17 4 7 12 13

19. After left rotation of ‘7’ 19 14 17 4 7 12 13

20. Question 2 In class we’ve seen an implementation of AVL tree where each node v has an extra field h, the height of the sub-tree rooted at v. The height can be used in order to balance the tree. -How many bits are required to store the height in a node? -Answer: For an AVL tree with n nodes, h=O(logn) thus requires O(loglogn) extra bits. 1.How can we reduce the number of the extra bits necessary for balancing the AVL tree? 2.Suggest an algorithm for computing the height of a given AVL tree given in the representation you suggested in 1. 20

21. Question 2 solution balance 1.Instead of a height field, which is redundant, each node will store 2 balance bits, calculated as the difference of heights between its right and left sub-trees. Two bits suffice because the difference can be one of the three: -1, 0, 1. (The leftmost bit represents the sign) The balance field should be updated on insert and delete operations, along the path to the root. 21

22. Question 2 solution balance 2.To compute the height of a tree, follow the path from the root to the deepest leaf by reading the balance field. If a sub tree is balanced to one side, the deepest leaf resides on that side. 22 CalcHeight(T) if T == null return -1 if T.balance == -1 or T.balance == 0 return 1 + CalcHeight( T.left ) else return 1 + CalcHeight( T.right )

23. Question 3 23

24. Question 3 solution 24

25. Question 3 solution 25

26. Question 3 solution Reminder: 26 TREE-SUCCESSOR(x) If x.right != NULL then return TREE-MINIMUM(x.right) y ← x.parent while y != NULL and x == y.right do x ← y y ← y.parent return y

27. Question 3 solution 27

28. Question 4 Suggest an efficient algorithm for sorting an array of numbers. Analyze its running time and required space. 28

29. Question 4 solution 29

30. Question 5 Suggest a data structure for storing integers that supports the following operations. 30 Init()Initialize the data structure.O(1) Insert(x)Insert x, if it is not present yet.O(log n) Delete(x)Delete x if it exists.O(log n) DeletePlace(i) Delete the element in the i th place (as determined by the order of insertion). O(log n) GetPlace(x) Return the place (which is determined by the order of insertion) of x. If x does not exist, return -1. O(log n)

31. Question 5 solution For example, for the following sequence of actions: Insert(3), Insert(5), Insert(11), Insert(4), Insert(7), Delete(5) GetPlace(7) returns 4, and DeletePlace(2) will delete 11. The solution We will use two AVL trees: T1 stores the elements by their key. T2 stores the elements by the order of insertion (using a running counter). There are pointers between the two trees connecting the nodes with the same key. 31

32. Question 5 solution 32