Problem of BST AVL Tree Binary Search is AVL trees are 4 not balanced. balanced. 44 2 3 Take this list of An AVL Tree is a 17 78 1 2 1 characters and binary search tree 32 50 88 form a tree such that for every 1 1 A B C D E F internal node v of T, 48 62 the heights of the children of v can differ by at most 1. An example of an AVL tree where the BST degenerates to a linked list heights are shown next to the nodes: AVL tree and 2-4 tree 1 AVL tree and 2-4 tree 2 Height of an AVL Tree Height of an AVL Tree (cont) Proposition: The height of an AVL tree T storing n keys is O(log n). Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2) n(h) > 2n(h-2) Justification: The easiest way to approach this problem n(h) > 4n(h-4) is to find n(h): the minimum number of internal nodes of n(h) > 8n(h-6) an AVL tree of height h. … We see that n(1) = 1 and n(2) = 2 n(h) > 2in(h-2i) For n = 3, an AVL tree of height h contains the root For any integer i such that h-2i ³ 1 Let h – 2i = 1, then i = (h – 1)/2 node, one AVL subtree of height n-1 and the other AVL Solving the base case we get: n(h) = 2 (h-1)/2 subtree of height n-2. Taking logarithms: h < 2log n(h) + 1 i.e. n(h) = 1 + n(h-1) + n(h-2) Thus the height of an AVL tree is O(log n) AVL tree and 2-4 tree 3 AVL tree and 2-4 tree 4 Insertion Insertion: rebalancing A binary search tree T is called balanced if for every node v, the height of v’s children differ by at most one. To rebalance the subtree rooted at z, we must Inserting a node into an AVL tree changes the heights of perform a restructuring some of the nodes in T. we rename x, y, and z to a, b, and c based on If an insertion causes T to become unbalanced, we travel the order of the nodes in an in-order traversal. up the tree from the newly created node until we find the first node x such that its grandparent z is unbalanced z is replaced by b, whose children are now a node. and c whose children, in turn, consist of the Since z became unbalanced by an insertion in the subtree four other subtrees formerly children of x, y, rooted at its child y , height(y ) = height(sibling(y )) + 2 and z. Now to rebalance... AVL tree and 2-4 tree 5 AVL tree and 2-4 tree 6 1 Insertion (contd.) Restructuring The four ways to rotate nodes in an AVL tree, graphically 5 44 represented 2 z 64 17 78 7 2 y -Single Rotations: 1 3 1 32 1 50 4 88 1 2 x 48 3 62 1 5 a = z single rotation b = y T ...balanced 54 3 b = y a = z c = x T c = x T 0 2 4 T T 44 0 3 T1 T T T T 3 4 x 3 0 1 2 2 T 17 62 2 unbalanced... y z 6 1 2 2 2 32 50 78 1 1 3 1 7 1 c = z single rotation b = y 5 b = y 48 54 88 a = x c = z a = x T 2 T3 T3 T0 T2 T2 T1 T0 T0 T 1 T3 T1 AVL tree and 2-4 tree 7 AVL tree and 2-4 tree 8 Restructuring (contd.) Restructure Algorithm function restructure(x): double rotations: Input: A node x of a binary search tree T that has both a parent y and a grandparent z Output: Tree T restructured by a rotation (either single or double) a = z double rotation b = x c = y a = z c = y involving nodes x, y, and z. b = x 1: Let (a, b, c) be an inorder listing of the nodes x, y, and z, and let (T0, T1, T2, T3) be an inorder listing of the the four subtrees of x, y, and T0 T2 T T T T T2 3 0 1 3 z. T 1 2. Replace the subtree rooted at z with a new subtree rooted at b 3. Let a be the left child of b and let T0, T1 be the left and right c = z double rotation b = x subtrees of a, respectively. a = y a = y c = z 4. Let c be the right child of b and let T2, T3 be the left and right b = x subtrees of c, respectively. T0 T2 T T T T 3 T2 3 1 0 T1 AVL tree and 2-4 tree 9 AVL tree and 2-4 tree 10 Cut/Link Restructure Algorithm Cut/Link Restructure Algorithm z 44 y Let’s go into a little more detail on this algorithm... 17 62 x Any tree that needs to be balanced can be grouped into 7 50 78 parts: x, y, z, and the 4 trees anchored at the children of T 0 48 54 88 those nodes (T0-T3) T 2 z T 1 T 3 44 y Make a new tree which is balanced and put the 7 parts 17 62 x from the old tree into the new tree so that the 50 78 numbering is still correct when we do an in-order- T0 48 54 88 traversal of the new tree. T 2 This works regardless of how the tree is originally T unbalanced. T 1 3 Let’s see how it works! AVL tree and 2-4 tree 11 AVL tree and 2-4 tree 12 2 Cut/Link Restructure Algorithm Cut/Link Restructure Algorithm Number the 7 parts by doing an in-order-traversal. (note that x,y, and z are now renamed based upon their order within the Now create an Array, numbered 1 to 7 (the 0th element can be ignored with minimal waste of space) traversal) 2 z (a) 1 44 y (b) 4 1 2 3 4 5 6 7 17 62 3 x (c) 6 •Cut() the 4 T trees and place them in their inorder rank in the array 50 78 T 0 48 54 5 88 7 T0 T1 T 2 T3 T 2 1 2 3 4 5 6 7 T 1 T 3 AVL tree and 2-4 tree 13 AVL tree and 2-4 tree 14 Cut/Link Restructure Algorithm Cut/Link Restructure Algorithm Now cut x,y, and z in that order (child,parent,grandparent) and place them in their inorderrank in the array. Link in ranks 2 (a) and 6 (c) as 4’s children. a b c b T0 T1 T 2 T3 4 44 62 78 62 a 1 2 3 4 5 6 7 2 c •Now we can re-link these subtrees to the main tree. 44 78 6 •Link in rank 4 (b) where the subtree’s root formerly 4 b 62 AVL tree and 2-4 tree 15 AVL tree and 2-4 tree 16 Cut/Link Restructure Algorithm Cut/Link Restructure algorithm Finally, link in ranks 1,3,5, and 7 as the children of 2 and 6. This algorithm for restructuring has the exact same effect as using the four rotation cases 4 y 62 discussed earlier. z x 2 44 Advantages: no case analysis, more elegant 78 6 5 7 Disadvantage: can be more code to write 3 50 88 17 Same time complexity T 48 54 2 T0 T3 T 1 • Now you have a balanced tree! AVL tree and 2-4 tree 17 AVL tree and 2-4 tree 18 3 Removal Removal (contd.) example of deletion from an AVL tree: We can easily see that performing a remove(w) can Oh no, unbalanced! 4 44 cause T to become unbalanced. y 1 3 17 62 x Let z be the first unbalanced node encountered 2 2 50 78 while traveling up the tree from w. Also, let y be the 1 1 0 1 T 0 48 54 88 child of z with the larger height, and let x be the T 2 32 child of y with the larger height. T 1 T 3 We can perform operation restructure(x) to restore y 4 62 z 3 x balance at the subtree rooted at z. 2 44 78 2 1 0 1 As this restructuring may upset the balance of 17 50 88 1 1 Whew, another node higher in the tree, we must continue 48 54 T 2 T 0 balanced! checking for balance until the root of T is reached T 3 AVL tree and 2-4 tree 19 AVL tree and 2-4 tree 20 Removal (contd.) AVL Trees - Data Structures example of deletion from an AVL tree: AVL trees can be implemented with a flag to z 4 indicate the balance state 44 y 1 3 Oh no, 17 x 62 2 2 unbalanced! 50 78 1 1 0 1 typedef enum {LeftHeavy, Balanced, T 0 48 54 88 RightHeavy} BalanceFactor; 32 T T T 1 2 3 typedef struct node { 4 x 50 BalanceFactor bf; z 2 y 3 44 62 void *item; 1 1 1 2 17 48 54 78 struct node *left, *right; 0 Whew, 1 88 balanced! } AVL_node; T 0 T 1 T 2 AVL tree and 2-4 tree 21 AVL tree and 2-4 tree 22 Multi-Way Search Tree A multi-way search tree is an ordered tree such that n Each internal node has at least two children and stores d -1 (2,4) Trees key-element items (ki, oi), where d is the number of children n For a node with children v1 v2 … vd storing keys k1 k2 … kd-1 w keys in the subtree of v1 are less than k1 w keys in the subtree of vi are between ki-1 and ki (i = 2, …, d - 1) 9 w keys in the subtree of vd are greater than kd-1 2 5 7 10 14 11 24 2 6 8 15 27 32 30 AVL tree and 2-4 tree 24 4 Multi-Way Inorder Traversal Multi-Way Searching We can extend the notion of inorder traversal from binary trees Similar to search in a binary search tree to multi-way search trees A each internal node with children v1 v2 … vd and keys k1 k2 … kd-1 Namely, we visit item (k, o) of node v between the recursive i i n k = ki (i = 1, …, d - 1): the search terminates successfully traversals of the subtrees of v rooted at children vi and vi + 1 n k < k1: we continue the search in child v1 An inorder traversal of a multi-way search tree visits the keys in n ki-1 < k < ki (i = 2, …, d - 1): we continue the search in child vi increasing order n k > kd-1: we continue the search in child vd Reaching an external node terminates the search unsuccessfully 11 24 Example: search for 30 8 12 2 6 8 15 27 32 11 24 2 4 6 10 14 18 30 2 6 8 15 27 32 1 3 5 7 9 11 13 16 19 30 15 17 AVL tree and 2-4 tree 25 AVL tree and 2-4 tree 26 (2,4) Tree Height of a (2,4) Tree Theorem: A (2,4) tree storing n items has height O(log n) A (2,4) tree (also called 2-4 tree or 2-3-4 tree) is a multi-way search with the following properties Proof: n Let h be the height of a (2,4) tree with n items n Node-Size Property: every internal node has at most four children i n Since there are at least 2 items at depth i = 0, … , h - 1 and no n Depth Property: all the external nodes have the same depth items at depth h, we have Depending on the number of children, an internal node of a n ³ 1 + 2 + 4 + … + 2h-1 = 2h - 1 (2,4) tree is called a 2-node, 3-node or 4-node n Thus, h £ log (n + 1) Searching in a (2,4) tree with n items takes O(log n) time 10 15 24 depth items 0 1 2 8 12 18 27 32 1 2 h-1 2h-1 h 0 AVL tree and 2-4 tree 27 AVL tree and 2-4 tree 28 Insertion Overflow and Split We insert a new item (k, o) at the parent v of the leaf reached by We handle an overflow at a 5-node v with a split operation: searching for k n let v1 … v5 be the children of v and k1 … k4 be the keys of v n We preserve the depth property but n node v is replaced nodes v' and v" n We may cause an overflow (i.e., node v may become a 5-node) w v' is a 3-node with keys k1 k2 and children v1 v2 v3 Example: inserting key 30 causes an overflow w v" is a 2-node with key k4 and children v4 v5 n key k is inserted into the parent u of v (a new root may be created) 10 15 24 3 v The overflow may propagate to the parent node u 2 8 12 18 27 32 35 u u 15 24 15 24 32 v v' v" 10 15 24 12 18 27 30 32 35 12 18 27 30 35 v 2 8 12 18 27 30 32 35 v1 v2 v3 v4 v5 v1 v2 v3 v4 v5 AVL tree and 2-4 tree 29 AVL tree and 2-4 tree 30 5 Analysis of Insertion Deletion function insertItem(k, o) Let T be a (2,4) tree We reduce deletion of an item to the case where the item is at the with n items node with leaf children 1.