Data Structure: Search Trees 2. Instructor: Prof. Young-guk Ha Dept. of Computer Science & Engineering

Transcription

1 Data Structure: Search Trees Instructor: Prof. Young-guk Ha Dept. of Computer Science & Engineering

2 Search Trees Tree data structures that can be used to implement a dictionary, especially an ordered dictionary Binary search trees AVL trees Multi-way search trees (2,4) trees Red-black trees 2

3 Multi-Way Search Trees MST

4 Multi-Way Search Trees A Multi-way Search Tree (MST) is an ordered tree s.t. eery internal node with d children stores d-1 entries (d 2) for a node with children 1, 2,, d storing keys k 1, k 2,, k d-1 keys in the node are ordered as k 1 k 2 k d-1 [eery key in the subtree of 1 ] k 1 k i-1 [eery key in the subtree of i ] k i (where 2 i d-1) [eery key in the subtree of d ] k d-1 The leaes store no entries and sere as placeholders just like BST k 1 k

5 Multi-Way Inorder Traersal We can extend the notion of the inorder traersal from binary trees to MSTs I.e., inorder traersal on an MST isits entry (k i, x i ) of node between the recursie traersals of subtrees rooted at children i and i+1 As in the case of BST, inorder traersal isits all the entries on the MST in an increasing order 1 2 k 1 k * Traersal order: [subtree of 1 ] k 1 [subtree of 2 ] k 2 [subtree of 3 ]

6 Multi-Way Searching Similar to the BST search To find an entry with key k, start from the root For each internal node with children 1,, d and keys k 1,, k d-1 if k is in node, the search terminates successfully if k < k 1, we continue the search in child 1 if k i-1 < k < k i (2 i d - 1), we continue the search in child i if k > k d-1, we continue the search in child d Reaching an external node terminates the search unsuccessfully Fail to search for 12 Successful search for 24 6

7 Implementation of MST MST implementation using a generic tree linked structure with secondary dictionaries D() storing entries of node key (k i ) alue parent [ Node ] 5 Dictionary D( i ) of entries k 1 k 1 D() [ Entry ] child ( i ) Values are not represented in the figure D( 1 ) 7 9 D( 2 ) 7

8 Analysis of Liked-Structure- Based MSTs The oerall space complexity is O(n), where n is the number of entries stored in an MST T Time complexity analysis Time spent at a node with d children O(log d), if D() is based on a search table O(d), if D() is based on a unsorted list Let d max denote the max number of children of any node in T and h the height of T, the search time is O(h*log d max ), if D() is based on a search table O(h*d max ), if D() is based on a unsorted list If d max is a constant, the running time is O(h), irrespectie of the implementation of the secondary dictionaries Thus, the primary efficiency goal for an MST is also to keep the height as small as possible 8

9 (2,4) Trees (2, 4)

10 Definition of (2,4) Tree A (2,4) Tree (also called tree) is a multi-way search tree with the following properties Size property: eery internal node has at most four children (d max = 4) i.e., D() search operations at any node are performed in O(1) time whether D() is sorted or not Depth property: all the external nodes hae the same depth enforces the important bound on the height Depending on the number of children (degree), each internal node of a (2,4) tree is called 2-node, 3-node or 4-node node 3-node 2-node

11 Height of a (2,4) Tree Theorem: height of a (2,4) tree storing n entries = O(log n) Proof Let h be the height of a (2,4) tree storing n entries Since there are at least 2 i entries at depth i = 0,, h - 1 and no entry at depth h (leaes), we hae n h-1 = 2 h - 1 By taking a log at the both sides, we get h log (n + 1), h = O(log n) Thus, searching a (2,4) tree with n entries takes O(log n) time Note search time of an MST is O(h*log d max ) and d max of a (2,4) tree is a constant (d max = 4) depth 0 1 # of entries h-1 h 2 h-1 (= 4)

12 Insertion Inserting an entry (k, x) into a (2,4) tree 1) Let denote a node that is the parent of the leaf z reached by unsuccessfully searching for k 2) Add a new child w to on the left of z 3) Insert a new entry (k, x) into the node between w and z E.g.) Insert key z w z 12

13 Insertion (cont.) Insertion preseres the depth property, but may iolate the size property E.g.) After inserting key 30, an oerflow occurs at node node becomes 5-node z * Oerflow: node becomes 5-node w z 13

14 Oerflow and Split We can handle an oerflow at 5-node with the split operation Let 1 5 be the children of and k 1 k 4 be the keys of 1) Node is replaced with nodes ' and " ' is a 3-node with keys k 1 k 2 and children " is a 2-node with key k 4 and children 4 5 2) Key k 3 is inserted into the parent u of Note the oerflow may propagate to the parent node u Then, split u If u is the root, create new root and split old root u u u ' 35 "

15 Sequential Insertion Example Initial tree insert(6) insert(12) insert(15) oerflow new root split oerflow split insert(3) insert(5) insert(10) insert(8) 15

16 Cascading Split Example insert(17) oerflow oerflow split new root split 16

17 Analysis of Insertion Let T be a (2,4) tree with n entries Tree T has O(log n) height Step 1 takes O(log n) time because we isit at most O(log n) nodes Step 2 takes O(1) time Step 3 takes O(log n) time because each split takes O(1) time and we perform at most O(log n) splits toward the root Thus, an insertion in a (2,4) tree takes O(log n) time Algorithm insert(k, o) 1) Search for key k to locate the insertion node 2) Add the new entry (k, x) at node 3) while isoerflow() // handling // oerflow if isroot() create a new empty root aboe split() parent() 17

18 Deletion If the entry to be deleted is at the node with leaf children Just remoe the entry and a leaf child Otherwise, if the node storing the entry to be deleted has no leaf children Replace the entry with its inorder successor (or, equialently predecessor) and then delete the replaced entry E.g., to delete key 24, replace it with 27 (inorder successor) replace

19 Underflow and Transfer Deleting an entry from a node may cause an underflow I.e., node becomes 1-node (with one child and no key) To handle an underflow at node with parent u, we need to consider the following two cases Case 1: one of adjacent siblings w of is a 3-node or a 4-node Perform the transfer operation (transfer a child of w to ia u) 1) Moe an entry from u to 2) Moe an entry from w to u Transfer operation does NOT propagate underflow u 4 9 u w 6 8 underflow w

20 Underflow and Fusion Case 2: all the adjacent siblings of are 2-nodes Perform the fusion operation 1) Merge with an adjacent sibling w 2) Moe an entry from u to the merged node ' as 3-node After a fusion, the underflow may propagate to parent u if u is a 2-node If u is the root, delete u Otherwise, perform fusion or transfer with u u 9 14 u 9 w underflow ' 20

21 Sequential Deletion Example remoe(4) transfer remoe(12) fusion remoe(13) 21

22 Propagating Fusion Example underflow propagated fusion remoe the root remoe(14) fusion 22

23 Analysis of Deletion Let T be a (2,4) tree with n entries Tree T has O(log n) height In a deletion operation We isit O(log n) nodes to locate the node with the entry to be deleted And then, we may handle an underflow with a series of O(log n) fusions, followed by at most one transfer Each fusion and transfer takes O(1) time Thus, deleting an entry from a (2,4) tree takes a total O(log n) time 23

24 Summarizing Performance of (2,4) Trees s is the number of entries returned 24

25 Red-Black Trees

26 Definition of Red-Black Tree A Red-black Tree is a binary search tree with each node colored red or black in a way that satisfies the following 4 properties Root Property: the root is black External Property: eery leaf is black Internal Property: the children of a red node are black Depth Property: all the leaes hae the same black depth* 9 Black depth = # of black ancestors

27 Red-Black Trees s. Other Search Trees In comparison with AVL trees and (2,4) trees Red-black trees hae the same logarithmic time performance But AVL and (2,4) trees may perform O(log n) physical restructurings while red-black trees performs O(1) restructurings I.e., restructuring in red-black trees does NOT propagate (cf. recoloring) A red-black tree can be represented as a (2,4) tree, ice ersa By merging eery red node into its parent This proides important intuition that will be used to perform restructuring or recoloring in red-black trees From red-black tree to (2,4) tree 27

28 From (2,4) to Red-Black Tree w 13 or 14 w w 6 8 z 28

29 Height of a Red-Black Tree Theorem Height of a red-black tree storing n entries = O(log n) Proof The red-black tree is at most twice as high as its associated (2,4) tree whose height is O(log n) Note one (2,4) tree node becomes two leels of red-black tree nodes at maximum Height of a red-black tree 2 * height of (2,4) tree = 2 * O(log n) = O(log n) The search algorithm of a red-black tree is the same as that of a binary search tree Thus, by the aboe theorem, searching in a red-black tree takes O(log n) time 29

30 A red-black tree can be implemented with a linked structure for binary trees Each node has an additional color indicator field The space requirement of a red-black tree storing n keys is O(n) Implementation of Red-Black Tree 6 8 B 6 B B color indicator 8 R B 30

31 Insertion To perform insert operation 1) Execute the insertion algorithm for binary search trees with the new node z 2) Color red the newly inserted node z unless it is the root We presere the root, external, and depth properties 3) Check the color of the parent node of z If the parent of z is black, we also presere the internal property and we are done If the parent of z is red, we iolated the internal property and need a reorganization of the tree this is called double red problem 31

32 Insertion Examples Example: an insertion of 4 done z 4 Example: an insertion of 4 causing a double red double red z 4 32

33 Remedying a Double Red Consider a double red with newly added child z and its parent (let w be the sibling of ) Case 1: If w is black The double red is an incorrect placement of keys in a 4-node Restructuring: we perform the 4-node realignment Case 2: If w is red The double red corresponds to an oerflow Recoloring: we perform the equialent of a split w 2 4 z 6 7 w 2 4 z

34 Restructuring A restructuring remedies a child-parent double red when the parent red node has a black sibling It is equialent to restoring the correct alignment of keys in a 4-node After the restructuring, the internal property is restored and the other properties are presered z w 2 4 z 6 7 tri-node restructuring w realignment

35 Restructuring (cont.) There are four restructuring configurations depending on whether the double red nodes are left or right children Same as the tri-node restructuring in AVL trees double rotation single rotation realignment 35

36 Recoloring A recoloring remedies a child-parent double red when the parent red node has a red sibling The parent and its sibling w become black and the grandparent u becomes red (unless it is the root) It is equialent to performing a split on a 5-node The double red iolation may propagate up after recoloring Because u may become red Need to check and continue remedy until the double red is finally resoled w 2 4 z u 6 7 recoloring w 2 u 4 z split

37 Insertion Examples insert(7) insert(12) after restructuring after recoloring insert(3) insert(5) insert(15) after restructuring insert(14) afrter recoloring insert(18) 37

38 Insertion Examples (cont.) after restructuring insert(16) Another double red after recoloring insert(17) after restructuring 38

39 Analysis of Insertion Algorithm insert(k, o) 1) We search for key k to locate the insertion node z 2) We add the new entry (k, o) at node z and color z red 3) while isdoublered(z) // remedy // double red if isblack(sibling(parent(z))) restructure(z) return else // if sibling(parent(z)) is red recolor(z) z parent(parent(z)) Recall that a red-black tree has O(log n) height Step 1 takes O(log n) time for a BST search Step 2 takes O(1) time Step 3 takes O(log n) time because we may perform O(log n) recolorings, each taking O(1) time, and at most one restructuring taking O(1) time Thus, an insertion in a red-black tree takes O(log n) time 39

40 Deletion To perform remoe operation 1) First execute the deletion algorithm for binary search trees 2) If the remoed node was red, we are done 3) Or if the remoed node was black, let be the remoed node, w the remoed external child of, and r the sibling of w (either external or internal) If r is red, we color r black and we are done If r is black, we will hae a double black problem, which iolates the depth property 40

41 Deletion Examples Example: deletion of r 7 w color r black 7 r black depth = 2 black depth = 2 Example: deletion of 8 causing a double black r w y double black r black depth = 1 black depth = 2 41

42 Remedying a Double Black Basically, a double black corresponds to an underflow from the iew of (2,4) trees The algorithm for remedying a double black node r with sibling y considers the following three cases Case 1: y is black and has a red child Perform a restructuring, equialent to a transfer of (2,4) trees, and we are done Case 2: y is black and its children are both black Perform a recoloring, equialent to a fusion of (2,4) trees Recoloring may propagate up the double black problem Case 3: y is red Perform an adjustment (actually a restructuring), equialent to a 3-node realignment in (2,4) trees After an adjustment either Case 1 or Case 2 needs to be applied according to the resulting configuration (note that, in this case, een if recoloring is applied, double-black is NOT propagated) 42

43 Restructuring after a Deletion Case 1: if y is black and has a red child x black depth -1 x 1-node r 1-node r transfer black depth +1 restructuring: (a) single or (b) double rotation r : either red or black 43

44 Recoloring after a Deletion Case 2: if y is black and its children are both black black depth -1 ` 2-node black depth -1 fusion recoloring black depth +1? fusion ` 1-node s? recoloring black depth -1 Recoloring propagates double black 44

45 Adjustment after a Deletion Case 3: if y is red black depth = 1 r realignment of a 3-node tri-node restructuring (adjustment)? r? s black depth = 1 * double-black is not resoled yet apply case 1 or 2 again according to the colors of node s and s s children (* note double black is not propagated after recoloring) 45

46 Deletion Examples remoe(12) remoe(3) after restructuring restructuring remoe(17) 46

47 Deletion Examples (cont.) remoe(18) recoloring remoe(15) remoe(16) adjustment after adjustment Double black is not resoled yet recoloring double black is resoled 47

48 Analysis of Deletion The deletion algorithm basically has the same procedure as the insertion algorithm A BST search takes O(log n) time A BST deletion takes O(1) time A reorganization procedure consisting of O(log n) recolorings, each taking O(1) time At most one adjustment (restructuring) taking O(1) time One additional restructuring or recoloring after the adjustment each taking O(1) time Thus, a deletion in a red-black tree also takes a total of O(log n) time 48

49 Summary of Red-Black Tree Reorganization Remedying double red after a insertion Red-black tree action (2,4) tree action Result Restructuring 4-node Realignment Double red remoed Recoloring Split Double red remoed or propagated up Remedying double black after a deletion Red-black tree action (2,4) tree action Result Restructuring Transfer Double black remoed Recoloring Adjustment Fusion 3-node Realignment Double black remoed or propagated up A restructuring or a recoloring follows 49

50 Performance of Red-Black Trees searches * s is the number of entries returned reorganization - O(1) restructurings - O(log n) recolorings Red-black, AVL, and (2,4) trees hae the same update operation (insertion and remoal) performance of O(log n) Howeer, red-black trees performs at most O(1) physical restructurings during the up-phase while AVL and (2,4) trees perform O(log n) restructurings 50