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Foundations of Data Structures Practical Session #7 AVL Trees 2

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

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AVL Tree example

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

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

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

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

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After left rotation of ’11’ Now balanced!

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

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

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Now balanced!

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Remove

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Unbalanced! Right rotate ’14’

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Balanced! Remove

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Remove 11 Replace it with the maximum in its left branch

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Remove

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Unbalanced! Required double rotatation

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After right rotation of ‘14’

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After left rotation of ‘7’

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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

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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

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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 )

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Question 3 23

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Question 3 solution 24

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Question 3 solution 25

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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

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Question 3 solution 27

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

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Question 4 solution 29

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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)

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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

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Question 5 solution 32

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