16. Binary Search Trees

Transcription

1 Dictionary imlementation 16. Binary Search Trees [Ottman/Widmayer, Ka..1, Cormen et al, Ka ] Hashing: imlementation of dictionaries with exected very fast access times. Disadvantages of hashing: linear access time in worst case. Some oerations not suorted at all: enumerate keys in increasing order next smallest key to given key 1 1 Trees Trees Trees are Generalized lists: nodes can have more than one successor Secial grahs: grahs consist of nodes and edges. A tree is a fully connected, directed, acyclic grah. Use Decision trees: hierarchic reresentation of decision rules syntax trees: arsing and traversing of exressions, e.g. in a comiler Code tress: reresentation of a code, e.g. morse alhabet, huffman code Search trees: allow efficient searching for an element by value 0 1

2 Examles Examles short start long / E T I A N M S U R W D K G O / 7.0 H V F U L A P I B X C Y Z Q Ö CH Morsealhabet Exression tree Nomenclature Binary Trees Wurzel A binary tree is either I inner node W E arent a leaf, i.e. an emty tree, or an inner leaf with two trees T l (left subtree) and T r (right subtree) as left and right successor. leaves K child Order of the tree: maximum number of child nodes, here: Height of the tree: maximum ath length root leaf (here: ) In each node v we store a key v.key and two nodes v.left and v.right to the roots of the left and right subtree. a leaf is reresented by the null-ointer left key right

3 Binary search tree A binary search tree is a binary tree that fulfils the search tree roerty: Every node v stores a key Keys in the left subtree v.left of v are smaller than v.key Key in the right subtree v.right of v are larger than v.key Searching Inut : Binary search tree with root r, key k Outut : Node v with v.key = k or null v r while v null do if k = v.key then return v else if k < v.key then v v.left else v v.right return null 1 Search (1) null Height of a tree Insertion of a key The height h(t ) of a tree T with root r is given by { 0 if r = null h(r) = 1 + max{h(r.left), h(r.right)} otherwise. The worst case run time of the search is thus O(h(T )) Insertion of the key k Search for k If successful search: outut error Of no success: insert the key at the leaf reached Insert () 1

4 Remove node Remove node Three cases ossible: Node has no children Node has one child Node has two children [Leaves do not count here] 1 Node has no children Simle case: relace node by leaf. 1 remove() Remove node Remove node Node has one child Also simle: relace node by single child. 1 remove() 1 Node has two children The following observation hels: the smallest key in the right subtree v.right (the symmetric successor of v) is smaller than all keys in v.right is greater than all keys in v.left and cannot have a left child. Solution: relace v by its symmetric successor. 1

5 By symmetry... Algorithm SymmetricSuccessor(v) Node has two children Also ossible: relace v by its symmetric redecessor. 1 Inut : Node v of a binary search tree. Outut : Symmetric successor of v w v.right x w.left while x null do w x x x.left return w Analysis Traversal ossibilities Deletion of an element v from a tree T requires O(h(T )) fundamental stes: Finding v has costs O(h(T )) If v has maximal one child unequal to nullthen removal takes O(1) stes Finding the symmetric successor n of v takes O(h(T )) stes. Removal and insertion of n takes O(1) stes. reorder: v, then T left (v), then T right (v).,,,,,,, 1 ostorder: T left (v), then T right (v), then v.,,,,, 1,, inorder: T left (v), then v, then T right (v).,,,,,,,

6 Further suorted oerations Degenerated search trees Min(T ): Read-out minimal value in O(h) ExtractMin(T ): Read-out and remove minimal value in O(h) List(T ): Outut the sorted list of elements Join(T 1, T ): Merge two trees with max(t 1 ) < min(t ) in O(n). 1 1 Insert,,,,,,1 ideally balanced 1 Insert,,,,,,1 linear list 1 Insert 1,,,,,, linear list Objective 17. AVL Trees Balanced Trees [Ottman/Widmayer, Ka..-..1, Cormen et al, Ka. Problem -] Searching, insertion and removal of a key in a tree generated from n keys inserted in random order takes exected number of stes O(log n). But worst case Θ(n) (degenerated tree). Goal: avoidance of degeneration. Artificial balancing of the tree for each udate-oeration of a tree. Balancing: guarantee that a tree with n nodes always has a height of O(log n). Adelson-Venskii and Landis (16): AVL-Trees 0 1

7 Balance of a node AVL Condition The height balance of a node v is defined as the height difference of its sub-trees T l (v) and T r (v) bal(v) := h(t r (v)) h(t l (v)) h l bal(v) T l (v) v T r (v) h r AVL Condition: for eacn node v of a tree bal(v) { 1, 0, 1} v h h + 1 T l (v) T r (v) h + (Counter-)Examles Number of Leaves AVL tree with height AVL tree with height No AVL tree 1. observation: a binary search tree with n keys rovides exactly n + 1 leaves. Simle induction argument.. observation: a lower bound of the number of leaves in a search tree with given height imlies an uer bound of the height of a search tree with given number of keys.

8 Lower bound of the leaves Lower bound of the leaves for h > Height of one subtree h 1. Height of the other subtree h. v h h 1 h Minimal number of leaves M(h) is AVL tree with height 1 has M(1) := leaves. M(h) = M(h 1) + M(h ) T l (v) T r (v) AVL tree with height has at least M() := leaves. Overal we have M(h) = F h+ with Fibonacci-numbers F 0 := 0, F 1 := 1, F n := F n 1 + F n for n > [Fibonacci Numbers: closed form] [Fibonacci Numbers: closed form] Closed form of the Fibonacci numbers: comutation via generation functions: 1 Power series aroach f(x) := F i x i For Fibonacci Numbers it holds that F 0 = 0, F 1 = 1, F i = F i 1 + F i i > 1. Therefore: f(x) = x + F i x i = x + F i 1 x i + F i x i = x + x = x + x i= i= F i 1 x i 1 + x i= F i x i + x = x + x f(x) + x f(x). i= F i x i i= F i x i

9 [Fibonacci Numbers: closed form] [Fibonacci Numbers: closed form] Thus: with the roots φ and ˆφ of 1 x x. f(x) (1 x x ) = x. x f(x) = 1 x x x f(x) = (1 φx) (1 ˆφx) φ = 1 + ˆφ = 1. 0 It holds that: Damit: (1 ˆφx) (1 φx) = x. f(x) = 1 (1 ˆφx) (1 φx) (1 φx) (1 ˆφx) = 1 ( 1 1 φx 1 ) 1 ˆφx 1 [Fibonacci Numbers: closed form] Power series of g a (x) = 1 1 a x (a R): 1 1 a x = a i x i. E.g. Taylor series of g a (x) at x = 0 or like this: Let G i x i a ower series of g. By the identity g a (x)(1 a x) = 1 it holds that 1 = G i x i a G i x i+1 = G 0 + (G i a G i 1 ) x i Thus G 0 = 1 and G i = a G i 1 G i = a i. i=1 [Fibonacci Numbers: closed form] 6 Fill in the ower series: f(x) = 1 ( 1 1 φx 1 ) ( 1 ˆφx = 1 φ i x i 1 = (φ i ˆφ i )x i ) ˆφ i x i Comarison of the coefficients with f(x) = F i x i yields F i = 1 (φ i ˆφ i ).

10 Fibonacci Numbers It holds that F i = 1 (φ i ˆφ i ) with roots φ, ˆφ of the equation x = x + 1 (golden ratio), thus φ = 1+, ˆφ = 1. Proof (induction). Immediate for i = 0, i = 1. Let i > : F i = F i 1 + F i = 1 (φ i 1 ˆφ i 1 ) + 1 (φ i ˆφ i ) = 1 (φ i 1 + φ i ) 1 ( ˆφ i 1 + ˆφ i ) = 1 φ i (φ + 1) 1 ˆφi ( ˆφ + 1) = 1 φ i (φ ) 1 ˆφi ( ˆφ ) = 1 (φ i ˆφ i ). Tree Height Because ˆφ < 1, overal we have ( M(h) Θ 1 + ) h Ω(1.61 h ) and thus h 1. log n + c. AVL tree is asymtotically not more than % higher than a erfectly balanced tree. Insertion Balance at Insertion Point Balance Kee the balance stored in each node Re-balance the tree in each udate-oeration New node n is inserted: Insert the node as for a search tree. Check the balance condition increasing from n to the root. n case 1: bal() = +1 case : bal() = 1 Finished in both cases because the subtree height did not change n 6 7

11 Balance at Insertion Point uin() - invariant case.1: bal() = 0 right n case.: bal() = 0, left n When uin() is called it holds that the subtree from is grown and bal() { 1, +1} Not finished in both case. Call of uin() uin() Assumtion: is left son of case 1: bal() = +1, done. 0 1 case : bal() = 0, uin() uin() Assumtion: is left son of 1 case : bal() = 1, In both cases the AVL-Condition holds for the subtree from This case is roblematic: adding n to the subtree from has violated the AVL-condition. Re-balance! 17 If is a right son: symmetric cases with exchange of +1 and 1 Two cases bal() = 1, bal() =

12 Rotationen Rotationen case 1.1 bal() = 1. 1 case 1.1 bal() = 1. 1 y 1 x 0 z 1 y 0 x 1 t h 1 t t 1 h 1 h rotation right t 1 t t h h 1 h 1 y 0 x +1 t 1 t h 1 t h h 1 h y h h 1 t h 1 double rotation left-right x t t 1 t t h h 1 h 1 h 1 h 1 h z 1 right son: bal() = bal() = +1, left rotation 1 right son: bal() = +1, bal() = 1, double rotation right left 6 6 Analysis Tree height: O(log n). Insertion like in binary search tree. Balancing via recursion from node to the root. Maximal ath lenght O(log n). Insertion in an AVL-tree rovides run time costs of O(log n). Deletion Case 1: Children of node n are both leaves Let be arent node of n. Other subtree has height h = 0, 1 or. h = 1: Adat bal(). h = 0: Adat bal(). Call uout(). h = : Rebalanciere des Teilbaumes. Call uout(). n h = 0, 1, h = 0, 1, 6 6

13 Deletion Deletion Case : one child k of node n is an inner node Relace n by k. uout(k) Case : both children of node n are inner nodes Relace n by symmetric successor. uout(k) n k Deletion of the symmetric successor is as in case 1 or. k uout() uout() Case (a).: bal() = +1. Let q be brother of (a)..1: bal(q) = 0. 0 Let be the arent node of. (a) left child of y +1 z 1 1 bal() = 1 bal() 0. uout() bal() = 0 bal() +1. bal() = +1 next slides. (b) right child of : Symmetric cases exchanging +1 and 1. x 0 q z 0 1 h 1 h 1 Left Rotate(y) x 0 1 y +1 h + 1 h + 1 h 1 h 1 h + 1 h (b)..1: bal() = 1, bal(q) = 1, Right rotation 6

14 uout() uout() Case (a).: bal() = +1. (a)..: bal(q) = Case (a).: bal() = +1. (a)..: bal(q) = 1. y +1 r z 0 y +1 r w 0 x 0 q z +1 1 h 1 h 1 h h + 1 Left Rotate(y) y 0 x 0 1 h 1 h 1 h h + 1 lus uout(r). x 0 q z 1 h 1 h 1 w 1 h Rotate right (z) left (y) x y h 1 h 1 h lus uout(r). z 1 (b)..: bal() = 1, bal(q) = +1, Right rotation+uout 70 (b)..: bal() = 1, bal(q) = 1, left-right rotation + uout 71 Conclusion AVL trees have worst-case asymtotic runtimes of O(log n) for searching, insertion and deletion of keys. Insertion and deletion is relatively involved and an overkill for really small roblems. 7