Data Structures and Algorithms Chapter 4: Trees (AVL Trees) Text: - PowerPoint PPT Presentation

CE 221 Data Structures and Algorithms Chapter 4: Trees (AVL Trees) Text: Read Weiss, § 4.4 Izmir University of Economics 1

AVL Trees • An AVL (Adelson-Velskii and Landis) tree is a binary search tree with a balance condition . It must be easy to maintain, and it ensures that the depth of the tree is O (log N ). • Idea 1: left and right subtrees are of the same height (not shallow). Izmir University of Economics 2

Balance Condition for AVL Trees • Idea 2: Every node must have left and right subtrees of the same height. The height of an empty tree is -1 (Only perfectly balanced trees with 2 k -1 nodes would satisfy). AVL Tree = BST with every node satisfying the property that the heights of left and right subtrees can differ only by one . The tree on the left is an AVL tree. The height info is kept for each node. Izmir University of Economics 3

The Height of an AVL Tree - I • For an AVL Tree with N nodes, the height is at most 1.44log( N +2)-0.328. • In practice it is slightly more than log N . • Example : An AVL tree of height 9 with fewest nodes (143). Izmir University of Economics 4

The Height of an AVL Tree - II • The minimum number of nodes, S ( h ), in an AVL tree of height h is given by S(0)=1, S(1)=2, S ( h ) = S ( h -1) + S ( h -2) +1 • Recall Fibonacci Numbers? ( F (0) = F (1) = 1, F ( n )= F ( n -1) + F ( n -2)) • Claim : S ( h ) = F ( h +2) - 1 for all h ≥ 0. • Proof : By induction. Base cases; S (0)=1= F (2)-1=2-1, S (1)=2= F (3)-1=3-1. Assume by inductive hypothesis that claim holds for all heights k ≤ h . Then, • S ( h +1) = S ( h ) + S ( h -1) + 1 = F ( h +2) -1 + F ( h +1) -1 + 1 (by inductive hypothesis) = F( h +3) – 1   h h       1 1 5 1 5 • We also know that         F ( h )     5  2 2  Izmir University of Economics 5

The Height of an AVL Tree - III    h h h 2          1 5 1 5 1   1 5 1            F ( h ) ( ) 2 N S h          2   2  5 2 5        h 2      h 1 1 5        1 1 5   2   N     F ( h ) 1     5  2  2 5          1 1 5           log( N 2 ) log 5 ( h 2 ) log S ( h ) F ( h 2 ) 1   2  2     h 2     1 5 1   log( N 2 ) log 5         S ( h ) 1 1 2 h           2 5   1 5 1 5         log 2 * log       2   2  h 2    1 1 5        1 . 44 log( N 2 ) 0 . 327 h S ( h ) 2    5  2  h O (log n ) • Thus, all the AVL tree operations can be performed in O (log N ) time. We will assume lazy deletions. Except possibly insertion (update all the balancing information and keep it balanced) Izmir University of Economics 6

AVL Tree Insertion • Example : Let’s try to insert 6 into the AVL tree below. This would destroy the AVL property of the tree. Then this property has to be restored before the insertion step is considered over. • It turns out that this can always be done with a simple modification to the tree known as rotation . After an insertion, only the nodes that are on the path from the insertion point to the root might have their balance altered. As we follow the path up to the root and update the balancing information there may exist nodes whose new balance violates the AVL condition. We will prove that our rebalancing scheme performed once at the deepest such node works . 6 Izmir University of Economics 7

AVL Tree Rotations • Let’s call the node to be balanced α . Since any node has at most 2 children, and a height imbalance requires that α ‘s 2 subtrees’ height differ by 2. There are four cases to be considered for a violation: 1) An insertion into left subtree of the left child of α . ( LL ) 2) An insertion into right subtree of the left child of α . ( LR ) 3) An insertion into left subtree of the right child of α . ( RL ) 4) An insertion into right subtree of the right child of α . ( RR ) • Cases 1 and 4 are mirror image symmetries with respect to α , as are Cases 2 and 3. Consequently; there are 2 basic cases. • Case I (LL, RR) (insertion occurs on the outside) is fixed by a single rotation . • Case II (RL, LR) (insertion occurs on the inside) is fixed by double rotation . Izmir University of Economics 8

Single Rotation (LL) • Let k 2 be the first node on the path up violating AVL balance property. Figure below is the only possible scenario that allows k 2 to satisfy the AVL property before the insertion but violate it afterwards. Subtree X has grown an extra level (2 levels deeper than Z now). Y cannot be at the same level as X ( k 2 then out of balance before insertion) and Y cannot be at the same level as Z (then k 1 would be the first to violate). Izmir University of Economics 9

Single Rotation (RR) • Note that in single rotation inorder traversal orders of the nodes are preserved. • The new height of the subtree is exactly the same as before. Thus no further updating of the nodes on the path to the root is needed. Izmir University of Economics 10

Single Rotation-Example I • AVL property destroyed by insertion of 6, then fixed by a single rotation. • BST node structure needs an additional field for height. Izmir University of Economics 11

Single Rotation-Example II • Start with an initially empty tree and insert items 1 through 7 sequentially. Dashed line joins the two nodes that are the subject of the rotation. Izmir University of Economics 12

Single Rotation-Example III Insert 6. Balance problem at the root. So a single rotation is performed. Finally, Insert 7 causing another rotation. Izmir University of Economics 13

Double Rotation (LR, RL) - I • The algorithm that works for cases 1 and 4 (LL, RR) does not work for cases 2 and 3 (LR, RL). The problem is that subtree Y is too deep, and a single rotation does not make it any less deep. • The fact that subtree Y has had an item inserted into it guarantees that it is nonempty. Assume it has a root and two subtrees. Izmir University of Economics 14

Double Rotation (LR) - II Below are 4 subtrees connected by 3 nodes. Note that exactly one of tree B or C is 2 levels deeper than D (unless all empty). To rebalance, k 3 cannot be root and a rotation between k 1 and k 3 was shown not to work. So the only alternative is to place k 2 as the new root. This forces k 1 to be k 2 ’s left child and k 3 to be its right child. It also completely determines the locations of all 4 subtrees. AVL balance property is now satisfied. Old height of the tree is restored; so, all the balancing and and height updating is complete. Izmir University of Economics 15

Double Rotation (RL) - III In both cases (LR and RL), the effect is the same as rotating between α ’s child and grandchild and then between α and its new child. Every double rotation can be modelled in terms of 2 single rotations. Inorder traversal orders are always preserved between k 1 , k 2 , and k 3 . Double RL = Single LL ( α ->right)+ Single RR ( α ) Double LR = Single RR ( α ->left)+ Single LL ( α ) Izmir University of Economics 16

Double Rotation Example - I • Continuing our example, suppose keys 8 through 15 are inserted in reverse order. Inserting 15 is easy but inserting 14 causes a height imbalance at node 7. The double rotation is an RL type and involves 7, 15, and 14. Izmir University of Economics 17

Double Rotation Example - II • insert 13: double rotation is RL that will involve 6, 14, and 7 and will restore the tree. Izmir University of Economics 18

Double Rotation Example - III • If 12 is now inserted, there is an imbalance at the root. Since 12 is not between 4 and 7, we know that the single rotation RR will work. Izmir University of Economics 19

Double Rotation Example - IV • Insert 11: single rotation LL; insert 10: single rotation LL; insert 9: single rotation LL; insert 8: without a rotation. Izmir University of Economics 20

Double Rotation Example - V • Insert 8½: double rotation LR. Nodes 8, 8½, 9 are involved. Izmir University of Economics 21

Implementation Issues - I • To insert a new node with key X into an AVL tree T , we recursively insert X into the appropriate subtree of T (let us call this T LR ). If the height of T LR does not change, then we are done. Otherwise, if a height imbalance appears in T , we do the appropriate single or double rotation depending on X and the keys in T and T LR , update the heights (making the connection from the rest of the tree above), and are done. Izmir University of Economics 22