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AVL Tree
An AVL tree is a self-balancing binary search tree that guarantees search, insertion, and deletion operations will take O(log n) time on average. It achieves this by ensuring the heights of the left and right subtrees of every node differ by at most one. When an insertion or deletion causes a height imbalance of two, rotations are performed to rebalance the tree.
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Introduction to algorithms and AVL trees. AVL trees are balanced binary search trees with height O(log N) for optimal operations.
Analysis of binary search tree operations. Best case O(log N) and worst case O(N) when unbalanced.
Discussion on tree balancing strategies including strict and self-adjusting methods. Introduction to algorithms for balancing like AVL trees.
Definition and properties of AVL trees, describing their balance conditions ensuring logarithmic height for efficient operations.
Explaining the height definition in AVL trees and bounding the minimum number of nodes in relation to height for balance.
Process for inserting nodes in AVL trees, checking balance factors, and correcting unbalance through rotations.
Methods for rebalancing trees after insertions, detailing four case scenarios for rotations needed to restore balance.
Overview of single and double rotations used in AVL trees during insertion to maintain balance.
Visual examples of AVL trees from various insertion scenarios to illustrate balancing.
Practice examples involving inserting elements in AVL trees with specified orders, reinforcing learning.
Deletion cases in AVL trees including leaf and two-children cases, along with the necessary rebalancing operations afterward.
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AVL Tree
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- 2. Balanced binary tree ●The disadvantage of a binary search tree is that its height can be as large as N-1 ● This means that the time needed to perform insertion and deletion and many other operations can be O(N) in the worst case ● We want a tree with small height ● A binary tree with N node has height at least (log N) ● Thus, our goal is to keep the height of a binary search tree O(log N) ● Such trees are called balanced binary search trees. Examples are AVL tree, red-black tree.
- 3. Binary Search Tree- Best Time ● All BST operations are O(h), where d is tree depth ● minimum d is 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) 2h log N
- 4. 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 depths of left and right subtree ■ Unbalanced degenerate tree
- 5. Balanced and unbalancedBST 4 2 5 1 3 1 5 2 4 3 7 6 4 2 6 5 71 3 Is this “balanced”?
- 6. Approaches to balancingtrees ● 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
- 7. Balancing Binary SearchTrees ● 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
- 8. AVL Tree is… ●Named after Adelson-Velskii and Landis ● the first dynamically balanced trees to be propose ● Binary search tree with balance condition in which the sub-trees of each node can differ by at most 1 in their height
- 9. Definition of abalanced tree ● Ensure the depth = O(log N) ● Take O(log N) time for searching, insertion, and deletion ● Every node must have left & right sub-trees of the same height
- 10. An AVL treehas the following properties: 1. Sub-trees of each node can differ by at most 1 in their height 2. Every sub-trees is an AVL tree
- 11. AVL tree? YES Each leftsub-tree has height 1 greater than each right sub-tree NO Left sub-tree has height 3, but right sub-tree has height 1
- 12. AVL tree Height ofa node ● The height of a leaf is 1. The height of a null pointer is zero. ● The height of an internal node is the maximum height of its children plus 1 Note that this definition of height is different from the one we defined previously (we defined the height of a leaf as zero previously).
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- 14. AVL Trees 12 8 16 410 2 6 14
- 15. AVL Tree -1 0 00 0 0 -1 00 1 -2 1 -10 0 AVL Tree AVLTree 0 Not an AVL Tree
- 16. Height of anAVL Tree ● Fact: The height of an AVL tree storing n keys is O(log n). ● Proof: Let us bound n(h): the minimum number of internal nodes of an AVL tree of height h. ● We easily see that n(1) = 1 and n(2) = 2 ● For n > 2, an AVL tree of height h contains the root node, one AVL subtree of height n-1 and another of height n-2. ● That is, n(h) = 1 + n(h-1) + n(h-2) ● Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2). So n(h) > 2n(h-2), n(h) > 4n(h-4), n(h) > 8n(n-6), … (by induction), n(h) > 2in(h-2i) ● Solving the base case we get: n(h) > 2 h/2-1 ● Taking logarithms: h < 2log n(h) +2 ● Thus the height of an AVL tree is O(log n) 3 4 n(1) n(2)
- 17. AVL - Goodbut 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
- 18. Height of anAVL Tree ● N(h) = minimum number of nodes in an AVL tree of height h. ● Basis ■ N(0) = 1, N(1) = 2 ● Induction ■ N(h) = N(h-1) + N(h-2) + 1 ● Solution (recall Fibonacci analysis) ■ N(h) > h ( 1.62) h-1 h-2 h
- 19. Height of anAVL 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(log n))
- 20. Insertion Insert 6 Imbalance at8 Perform rotation with 7
- 21. Deletion Delete 4 Imbalance at3 Perform rotation with 2 Imbalance at 5 Perform rotation with 8
- 22. Key Points ● AVLtree remain balanced by applying rotations, therefore it guarantees O(log N) search time in a dynamic environment ● Tree can be re-balanced in at most O(log N) time
- 23. Searching AVL Trees ●Searching an AVL tree is exactly the same as searching a regular binary tree ■ all descendants to the right of a node are greater than the node ■ all descendants to the left of a node are less than the node
- 24. Inserting in AVLTree ● Insertion is similar to regular binary tree ■ keep going left (or right) in the tree until a null child is reached ■ insert a new node in this position ○ an inserted node is always a leaf to start with ● Major difference from binary tree ■ must check if any of the sub-trees in the tree have become too unbalanced ○ search from inserted node to root looking for any node with a balance factor of 2
- 25. Inserting in AVLTree ● A few points about tree inserts ■ the insert will be done recursively ■ the insert call will return true if the height of the sub-tree has changed ○ since we are doing an insert, the height of the sub-tree can only increase ■ if insert() returns true, balance factor of current node needs to be adjusted ○ balance factor = height(right) – height(left) left sub-tree increases, balance factor decreases by 1 right sub-tree increases, balance factor increases by 1 ■ if balance factor equals 2 for any node, the sub- tree must be rebalanced
- 26. Inserting in AVLTree M(-1) insert(V)E(1) J(0) P(0) M(0) E(1) J(0) P(1) V(0) M(-1) insert(L)E(1) J(0) P(0) M(-2) E(-2) J(1) P(0) L(0) This tree needs to be fixed!
- 27. Re-Balancing a Tree ●To check if a tree needs to be rebalanced ■ start at the parent of the inserted node and journey up the tree to the root ○ if a node’s balance factor becomes 2 need to do a rotation in the sub-tree rooted at the node ○ once sub-tree has been re-balanced, guaranteed that the rest of the tree is balanced as well can just return false from the insert() method ■ 4 possible cases for re-balancing ○ only 2 of them need to be considered other 2 are identical but in the opposite direction
- 28. Let the nodethat 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. Insertions in AVL Trees
- 29. j k X Y Z Consider avalid AVL subtree AVL Insertion: Outside Case h h h
- 30. j k X Y Z Inserting into X destroysthe AVL property at node j AVL Insertion: Outside Case h h+1 h
- 31. j k X Y Z Do a “rightrotation” AVL Insertion: Outside Case h h+1 h
- 32. j k X Y Z Do a “rightrotation” Single right rotation h h+1 h
- 33. j k X Y Z “Rightrotation” done! (“Left rotation” is mirror symmetric) Outside Case Completed AVL property has been restored! h h+1 h
- 34. j k X Y Z AVL Insertion:Inside Case Consider a valid AVL subtree h hh
- 35. Inserting into Y destroysthe AVL property at node j j k X Y Z AVL Insertion: Inside Case Does “right rotation” restore balance? h h+1h
- 36. j k X Y Z “Right rotation” does notrestore balance… now k is out of balance AVL Insertion: Inside Case h h+1 h
- 37. Consider the structure ofsubtree Y… j k X Y Z AVL Insertion: Inside Case h h+1h
- 38. j k X V Z W i Y = nodei and subtrees V and W AVL Insertion: Inside Case h h+1h h or h-1
- 39. j k X V Z W i AVL Insertion: InsideCase We will do a left-right “double rotation” . . .
- 40. j k X V Z W i Double rotation: first rotation left rotation complete
- 41. j k X V Z W i Double rotation: second rotation Now do a right rotation
- 42. jk X V ZW i Doublerotation : second rotation right rotation complete Balance has been restored hh h or h-1
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- 49. 5/22/2012 3 Example ● Insert 3into the AVL tree 11 8 20 4 16 27 8 8 11 4 20 3 16 27
- 50. 5/22/2012 Example ● Insert 5into the AVL tree 5 11 8 20 4 16 27 8 11 5 20 4 16 27 8
- 51. 5/22/2012 AVL Trees: Exercise ●Insertion order: ■ 10, 85, 15, 70, 20, 60, 30, 50, 65, 80, 90, 40, 5, 55
- 52. 5/22/2012 Deletion X inAVL Trees ● Deletion: ■ Case 1: if X is a leaf, delete X ■ Case 2: if X has 1 child, use it to replace X ■ Case 3: if X has 2 children, replace X with its inorder predecessor (and recursively delete it) ● Rebalancing
- 53. 5/22/2012 Delete 55 (case1) 60 20 70 10 40 65 85 5 15 30 50 80 90 55
- 54. 5/22/2012 Delete 55 (case1) 60 20 70 10 40 65 85 5 15 30 50 80 90 55
- 55. 5/22/2012 Delete 50 (case2) 60 20 70 10 40 65 85 5 15 30 50 80 90 55
- 56. 5/22/2012 Delete 50 (case2) 60 20 70 10 40 65 85 5 15 30 50 80 90 55
- 57. 5/22/2012 Delete 60 (case3) 60 20 70 10 40 65 85 5 15 30 50 80 90 55 prev
- 58. 5/22/2012 Delete 60 (case3) 55 20 70 10 40 65 85 5 15 30 50 80 90
- 59. 5/22/2012 Delete 55 (case3) 55 20 70 10 40 65 85 5 15 30 50 80 90 prev
- 60. 5/22/2012 Delete 55 (case3) 50 20 70 10 40 65 85 5 15 30 80 90
- 61. 5/22/2012 Delete 50 (case3) 50 20 70 10 40 65 85 5 15 30 80 90 prev
- 62. 5/22/2012 Delete 50 (case3) 40 20 70 10 30 65 85 5 15 80 90
- 63. 5/22/2012 Delete 40 (case3) 40 20 70 10 30 65 85 5 15 80 90 prev
- 64. 5/22/2012 Delete 40 :Rebalancing 30 20 70 10 65 85 5 15 80 90 Case ?
- 65. 5/22/2012 Delete 40: afterrebalancing 30 7010 20 65 855 15 80 90 Single rotation is preferred!
- 66. 5/22/2012 AVL Tree: analysis ●The depth of AVL Trees is at most logarithmic. ● So, all of the operations on AVL trees are also logarithmic. ● The worst-case height is at most 44 percent more than the minimum possible for binary trees.