Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "Foundations of Data Structures Practical Session #7 AVL Trees 2."— Presentation transcript:

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

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

5 right A single right rotation of ’11’ is executed to rebalance the tree: Insert

6 Now insert The sub-tree of 11 is unbalanced. Double rotation: right and then left.

7 After right rotation of ’13’ Now left rotate ’11’

8 After left rotation of ’11’ Now balanced!

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

10 After right rotation of ’12’ Now left rotate ‘7’

11 Now balanced!

12 Remove

13 Unbalanced! Right rotate ’14’

14 Balanced! Remove

15 Remove 11 Replace it with the maximum in its left branch

16 Remove

17 Unbalanced! Required double rotatation

18 After right rotation of ‘14’

19 After left rotation of ‘7’

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


Download ppt "Foundations of Data Structures Practical Session #7 AVL Trees 2."

Similar presentations


Ads by Google