Lecture3

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Lecture3

  1. 1. Advance Data Structures Lecture 3 AVL Trees
  2. 2. Binary Search Tree - Best Time • All BST operations are O(h), where h is tree height • minimum h is h log 2N for a binary tree with N nodes › What is the best case tree? › What is the worst case tree? • So, best case running time of BST operations is O(log N) 2
  3. 3. Binary Search Tree - Worst Time • Worst case running time is O(N) › What happens when you Insert elements in ascending order? • Insert: 2, 4, 6, 8, 10, 12 into an empty BST › Problem: Lack of “balance”: • compare height of left and right subtree › Unbalanced degenerate tree 3
  4. 4. Balanced and unbalanced BST 1 4 2 2 5 3 1 4 4 6 3 Is this “balanced”? 5 2 1 3 5 6 7 7 4
  5. 5. Approaches to balancing trees • Don't balance › May end up with some nodes very deep • Strict balance › The tree must always be balanced perfectly • Pretty good balance › Only allow a little out of balance • Adjust on access › Self-adjusting 5
  6. 6. Balancing Binary Search Trees • Many algorithms exist for keeping binary search trees balanced › Adelson-Velskii and Landis (AVL) trees (height-balanced trees) › Splay trees and other self-adjusting trees › B-trees and other multiway search trees 6
  7. 7. Perfect Balance • Want a complete tree after every operation › tree is full except possibly in the lower right • This is expensive › For example, insert 2 in the tree on the left and then rebuild as a complete tree 6 5 4 1 9 5 8 Insert 2 & complete tree 1 2 8 4 6 9 7
  8. 8. AVL - Good but not Perfect Balance • AVL trees are height-balanced binary search trees • Balance factor of a node › height(left subtree) - height(right subtree) • An AVL tree has balance factor calculated at every node › For every node, heights of left and right subtree can differ by no more than 1 › Store current heights in each node 8
  9. 9. Height of an AVL Tree • N(h) = minimum number of nodes in an AVL tree of height h. • Basis › N(0) = 1, N(1) = 2 • Induction h › N(h) = N(h-1) + N(h-2) + 1 • Solution (recall Fibonacci analysis) › N(h) > h ( 1.62) h-1 h-2 9
  10. 10. Height of an AVL Tree • N(h) > h ( 1.62) • Suppose we have n nodes in an AVL tree of height h. › n > N(h) (because N(h) was the minimum) › n > h hence log n > h (relatively well balanced tree!!) › h < 1.44 log2n (i.e., Find takes O(logn)) 10
  11. 11. Node Heights Tree A (AVL) height=2 BF=1-0=1 Tree B (AVL) 2 6 6 1 0 1 1 4 9 4 9 0 0 0 0 0 1 5 1 5 8 height of node = h balance factor = hleft-hright empty height = -1 11
  12. 12. Node Heights after Insert 7 Tree A (AVL) 2 Tree B (not AVL) balance factor 1-(-1) = 2 3 6 6 1 1 1 2 4 9 4 9 0 0 0 0 0 1 1 5 7 1 5 -1 8 0 height of node = h balance factor = hleft-hright empty height = -1 7 12
  13. 13. Insert and Rotation in AVL Trees • Insert operation may cause balance factor to become 2 or –2 for some node › only nodes on the path from insertion point to root node have possibly changed in height › So after the Insert, go back up to the root node by node, updating heights › If a new balance factor (the difference hlefthright) is 2 or –2, adjust tree by rotation around the node 13
  14. 14. Single Rotation in an AVL Tree 2 2 6 6 1 2 1 1 4 9 4 8 0 0 1 0 0 0 0 1 5 8 1 5 7 9 0 7 14
  15. 15. Insertions in AVL Trees Let the node that needs rebalancing be . There are 4 cases: Outside Cases (require single rotation) : 1. Insertion into left subtree of left child of . 2. Insertion into right subtree of right child of . Inside Cases (require double rotation) : 3. Insertion into right subtree of left child of . 4. Insertion into left subtree of right child of . The rebalancing is performed through four separate rotation algorithms. 15
  16. 16. AVL Insertion: Outside Case Consider a valid AVL subtree j k h h h X Z Y 16
  17. 17. AVL Insertion: Outside Case j k h+1 Inserting into X destroys the AVL property at node j h h Z Y X 17
  18. 18. AVL Insertion: Outside Case j k h+1 Do a “right rotation” h h Z Y X 18
  19. 19. Single right rotation j k h+1 Do a “right rotation” h h Z Y X 19
  20. 20. Outside Case Completed “Right rotation” done! (“Left rotation” is mirror symmetric) k j h+1 h h X Y Z AVL property has been restored! 20
  21. 21. AVL Insertion: Inside Case Consider a valid AVL subtree j k h X h h Z Y 21
  22. 22. AVL Insertion: Inside Case Inserting into Y destroys the AVL property at node j j k h h X Does “right rotation” restore balance? h+1 Z Y 22
  23. 23. AVL Insertion: Inside Case k j h X “Right rotation” does not restore balance… now k is out of balance h h+1 Z Y 23
  24. 24. AVL Insertion: Inside Case Consider the structure of subtree Y… j k h h X h+1 Z Y 24
  25. 25. AVL Insertion: Inside Case j Y = node i and subtrees V and W k h i h X h+1 Z h or h-1 V W 25
  26. 26. AVL Insertion: Inside Case j We will do a left-right “double rotation” . . . k Z i X V W 26
  27. 27. Double rotation : first rotation j left rotation complete i Z k W X V 27
  28. 28. Double rotation : second rotation j Now do a right rotation i Z k W X V 28
  29. 29. Double rotation : second rotation right rotation complete Balance has been restored i j k h h h or h-1 X V W Z 29
  30. 30. Implementation balance (1,0,-1) key left right No need to keep the height; just the difference in height, i.e. the balance factor; this has to be modified on the path of insertion even if you don’t perform rotations Once you have performed a rotation (single or double) you won’t need to go back up the tree 30
  31. 31. Single Rotation RotateFromRight(n : reference node pointer) { p : node pointer; p := n.right; n n.right := p.left; p.left := n; n := p } You also need to modify the heights or balance factors of n and p X Insert Y Z 31
  32. 32. Double Rotation • Implement Double Rotation in two lines. DoubleRotateFromRight(n : reference node pointer) { ???? n } X Z V W 32
  33. 33. Insertion in AVL Trees • Insert at the leaf (as for all BST) › only nodes on the path from insertion point to root node have possibly changed in height › So after the Insert, go back up to the root node by node, updating heights › If a new balance factor (the difference hlefthright) is 2 or –2, adjust tree by rotation around the node 33
  34. 34. Insert in BST Insert(T : reference tree pointer, x : element) : integer { if T = null then T := new tree; T.data := x; return 1;//the links to //children are null case T.data = x : return 0; //Duplicate do nothing T.data > x : return Insert(T.left, x); T.data < x : return Insert(T.right, x); endcase } 34
  35. 35. Insert in AVL trees Insert(T : reference tree pointer, x : element) : { if T = null then {T := new tree; T.data := x; height := 0; return;} case T.data = x : return ; //Duplicate do nothing T.data > x : Insert(T.left, x); if ((height(T.left)- height(T.right)) = 2){ if (T.left.data > x ) then //outside case T = RotatefromLeft (T); else //inside case T = DoubleRotatefromLeft (T);} T.data < x : Insert(T.right, x); code similar to the left case Endcase T.height := max(height(T.left),height(T.right)) +1; return; } 35
  36. 36. Example of Insertions in an AVL Tree 2 20 0 10 Insert 5, 40 1 30 0 25 0 35 36
  37. 37. Example of Insertions in an AVL Tree 2 3 20 1 1 1 10 30 20 10 0 0 5 25 0 35 0 5 2 30 0 1 25 35 0 Now Insert 45 40 37
  38. 38. Single rotation (outside case) 3 3 20 1 2 1 10 30 20 10 0 0 5 25 2 35 0 5 2 30 0 40 1 25 0 Imbalance 35 1 40 0 45 0 45 Now Insert 34 38
  39. 39. Double rotation (inside case) 3 3 20 1 5 1 10 0 3 30 20 10 0 2 Imbalance 25 35 0 40 2 1 5 40 1 30 0 1 35 Insertion of 34 0 45 0 0 25 34 45 34 39
  40. 40. Extended Example Insert 3,2,1,4,5,6,7, 16,15,14
  41. 41. AVL Tree Deletion • Similar but more complex than insertion › Rotations and double rotations needed to rebalance › Imbalance may propagate upward so that many rotations may be needed. 41
  42. 42. Deletion of a Node • Deletion of a node x from an AVL tree requires the same basic ideas, including single and double rotations, that are used for insertion. • With each node of the AVL tree is associated a balance factor that is left high, equal or right high according, respectively, as the left subtree has height greater than, equal to, or less than that of the right subtree. 42
  43. 43. Method 1. Reduce the problem to the case when the node x to be deleted has at most one child. › › If x has two children replace it with its immediate predecessor y under inorder traversal (the immediate successor would be just as good) Delete y from its original position, by proceeding as follows, using y in place of x in each of the following steps. 43
  44. 44. Method (cont.) 2. Delete the node x from the tree. › We’ll trace the effects of this change on height through all the nodes on the path from x back to the root. › The action to be taken at each node depends on • • balance factor of the node sometimes the balance factor of a child of the node. 44
  45. 45. Rebalancing the tree Let p be the node whose balance factor becomes 2 or -2. › Rotation is needed. › Let q be the root of the taller subtree of p. We have three cases according to the balance factor of q. 45
  46. 46. Case a Case a: The balance factor of q is 0. › Apply a single rotation › Height of p is unchanged. height unchanged p 1 -1 0 h-1 deleted q q p -1 T1 h h T2 h T3 h-1 T1 h T3 T2 46
  47. 47. Case b Case b: The balance factor of q is the same as that of the previous balance factor of p. › Apply a single rotation › Set the balance factors of p and q to 0 height reduced › Height of the subtree reduced . p 0 -1 -1 h-1 q T1 h-1 deleted p T2 h T3 h-1 0 h h-1 T1 q T3 T2 47
  48. 48. Case c Case c: The previous balance factors of p and balance factor of q are opposite. › Apply a double rotation › set the balance factors of the new root to 0 › Height of subtree is reduced. height reduced -1 p 0 1 h-1 q r p q r T1 h-1 T2 h-1 or h-2 T4 h-1 T3 T1 T2 h-1 or h-2 T4 T3 h-1 48
  49. 49. Delete 55 60 20 70 10 5 40 15 30 65 85 80 50 55 90
  50. 50. Delete 55 60 20 70 10 5 40 15 30 65 85 80 50 55 90
  51. 51. Delete 50 60 20 70 10 5 40 15 30 65 85 80 50 55 90
  52. 52. Delete 50 60 20 70 10 5 40 15 30 65 50 55 85 80 90
  53. 53. Delete 60 60 20 70 10 5 40 15 30 65 50 prev 55 85 80 90
  54. 54. Delete 60 55 20 70 10 5 40 15 30 65 50 85 80 90
  55. 55. Delete 40 40 20 10 5 70 30 15 65 85 80 90
  56. 56. Delete 40 40 20 10 5 30 15 70 prev 65 85 80 90
  57. 57. Delete 40 : Rebalancing 30 20 10 5 Case ? 15 70 65 85 80 90
  58. 58. Delete 40: after rebalancing 30 10 70 20 5 15 Single rotation is preferred! 65 85 80 90
  59. 59. Delete 20 30 20 10 55 25 15 45 60 50 59 65 57 59
  60. 60. Delete 20 30 25 10 55 45 15 60 50 59 65 57 60
  61. 61. Delete 20 30 15 10 55 25 45 60 50 59 65 57 61
  62. 62. Delete 20 55 30 15 10 60 59 45 25 50 65 57 62
  63. 63. Pros and Cons of AVL Trees Arguments for AVL trees: 1. Search is O(log N) since AVL trees are always balanced. 2. Insertion and deletions are also O(logn) 3. The height balancing adds no more than a constant factor to the speed of insertion. Arguments against using AVL trees: 1. Difficult to program & debug; more space for balance factor. 2. Asymptotically faster but rebalancing costs time. 3. Most large searches are done in database systems on disk and use other structures (e.g. B-trees). 4. May be OK to have O(N) for a single operation if total run time for many consecutive operations is fast (e.g. Splay trees). 63
  64. 64. Double Rotation Solution DoubleRotateFromRight(n : reference node pointer) { RotateFromLeft(n.right); n RotateFromRight(n); } X Z V W 64

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