TREES. General Binary Trees The Search Tree ADT Binary Search Trees AVL Trees Threaded trees Splay Trees B-Trees. UNIT -II

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1 UNIT -II TREES General Binary Trees Te Searc Tree DT Binary Searc Trees VL Trees Treaded trees Splay Trees B-Trees. 2MRKS Q& 1. Define Tree tree is a data structure, wic represents ierarcical relationsip Between individual data item. tree is a finite set one or more nodes suc tat Tere is a specially designated node called root. Te remaining nodes are partitioned into n>= 0 disjoint sets a, b, c, d. n Were a, b, c, d are called te s te root tree is sown in te following fig. T L R a b c d 2.Define Non-linear data structure. (OR) Give two examples non-linear data structures, wic are widely used. (pr/may2004) Data structure, wic is capable expressing more complex relationsip tan tat pysical adjacent, is called non-linear data structure. Example: Trees, grap, lists 1

2 3.Wat is meant by Binary tree? (pr/may2004) binary tree is a finite set nodes wic is eiter empty or consists a root and two disjoint binary trees called te left sub-tree and rigt sub-tree.(or)in an m-ary tree, if m=2, ten it is a binary tre Eac node as not more tan 2 cild nodes, eac wic may be a Leaf node. pplication: Binary tree is used in data processing. File index scemes Hierarcical database management systems. 4. Wat are te various operations performed on binary tree using linked representation? Te operation binary tree tat use linked list representation are, Insertion Insert a given item into a tree. Traversal Processing te nodes te tree one by one. Searc Searc for te specified item. Copying To obtain te exact copy te given tree. Deletion Delete an item from te tree. 5. Wat are te two metods binary tree implementation? Two metods to implement binary tree are, Linear representation. Linked representation. 6. Wat are te different types tree traversal tecnique? (pr/may2004) Te different types tree traversal tecnique are, Pre-order traversal yields prefix form expression. In-order traversal - yields Infix form expression. Post- order traversal - yields Postfix form expression. 7. Define Pre-order traversal. Pre-order traversal entains te following steps: Process te root node. Process te left. Process te rigt. E.g. + B O/P: +B 2

3 8.Define Post- order traversal. Post- order traversal entrains te following steps: Process te left. Process te rigt. Process te root node. E.g. + O/P: B+ B 9. Define In-order traversal. In-order traversal entains te following steps: Process te left. Process te root node. Process te rigt. E.g. + O/P: +B B 10.Define complete binary tree. binary tree is said to be a complete binary tree, if all its level, except possibly te last level, ave te maximum number possible nodes, all te nodes at te last level appear as far left as possible. T L R B C 11.Define Full binary tree. binary tree is a full binary tree, if it contains maximum possible number nodes in all level. 3

4 T L R B C D 12. Sow te postorder traversal te following binary searc tree: 13. For te following code fragment, use summations to calculate te exact number times te inner for loop executes, and give a tigt upper bound (Big-O analysis) te running time. Include an evaluation te summations. for ( int i = 0; i < n * n; i++ ) for ( int j = 0; j < i; j++ ) sum ++; 14. Te following algoritm determines if an integer i exists suc tat [i] = i in an array increasing integers; it returns i if it exists and -1 if not. Give te recurrence relation tat represents te running time te algoritm. Solve te recurrence relation using one te metods sown in class and give a tigt upper bound (Big-O analysis) te running time. Do NOT include te pro tat te solution to your recurrence relation is correct. int findmatc (, p, q ) if p < q m ( p + q ) / 2; 4

5 if [ m ] = m return m; else if m < [ m ] return findmatc(, p, m-1 ); else return findmatc(, m+1, q ); return -1; 15. Wat is meant by eigt balanced trees or VL trees? tree is said to be eigt balanced if all its nodes ave a balance factor 1, 0 or 1.Heigt balancing, attempts to maintain trees tat posses te ordering property in a form close to fullness. VL dulation VelsKii and Landis. 16. Wat is a balance factor in VL trees? Balance factor a node is defined to be te difference between te eigts te node s left and te eigt te node s rigt. 17. Define VL rotation. Manipulation tree pointers is centered at te pivot node to bring te tree into eigt balance. Te visual effect tis pointer manipulation is to rotate te wose root is te pivot node. Tis operation is referred as VL rotation. 18. Wat are te four-different cases pointer manipulations done in VL rotation? Te four-different cases pointer manipulations are, Insertion occurred in te left te left cild te pivot node. Insertion occurred in te rigt te rigt cild te pivot node. Insertion occurred in te rigt te left cild te pivot node. Insertion occurred in te left te rigt cild te pivot nod 19. Define Binary Searc tree. Binary Searc tree t is a binary tree; eiter it is empty or eac node in te tree contains an identifier and ll identifier in te left t are less (numerically or alpanumerically) tan te identifier in te root Node +. ll identifier in te rigt t are greater tan te identifier in te root Node +. 5

6 Te left and rigt s t are also binary searc trees. 20. Write all te Tree Terminologies? 1. Root: Tis is te unique node in te tree to wic furter sub-trees are attaced. In te Fig 1 is te root node. 2. degree te node Te total number sub-trees attaced to tat node is called te degree te node. For node te degree is 3, for node E te dgree is Leaves Tese are te terminal nodes te tree. Te nodes wit degree 0 are always te leaves. Here in our example E, F, C, G, H are te leaf nodes. B C D E F G Fig 1 4. Parent node Te node wic is aving furter sub-brances is called te parent node tose sub-brances. In te above fig te node B is parent node E, F, G nodes. nd E, F, G are called te cildren te parent B. 5. Siblings Nodes wit te same parent are siblings. Here node B is parent E, F, G called siblings. 21. Wat is meant by Level te tree? Te root node is always considered at level zero, ten its adjacent cildren are supposed to be at level 1 and so on. To understand te concept level, Fig 1 te node is at level 0, te nodes B, C, D are at level 1. te nodes E, F, G, H are at level 2. H 22. Wat is te Heigt te Tree? B C D E F G H 6

7 Te maximum level is te eigt te tree. Here te eigt te tree is 4. Te oter terminology used for eigt te tree is dept te tree. B C D E F G H I 23. Traverse te given tree using Inorder, Preorder and Postorder traversals. Given tree: B C D E F G H I J Inorder : D H B E F C I G J Preorder: B D H E C F G I J Postorder: H D E B F I J G C 24. Wat are te operations preformed in te Binary Searc Trees? Binary searc trees store collections items tat can be ordered, suc as integers. Binary searc trees support te following standard operations. searc(x): determines is an item x contained in te tree (and if so returns te item). 7

8 insert(x): adds item x to te collection stored in te tree, if it is not already tere. delete(x): removes item x from te collection stored in te tree. Data structures supporting tese operations are called Dictionaries. simple implementation is to store te items in sorted order in an array. Searc(x) is simply a binary searc. However, insert(x) and delete(x) are inefficient as, in general, tey may require many items to be sifted. 25. Draw a binary Tree for te expression: * B - (C + D) * (P / Q) - * * B + / C D P Q 26. Wat is meant by ancestor and descendant? B C D E F G H I J Wile displaying te tree, if some particular node occurs previous to some oter node ten tat node is called te ancestor te oter node. In our example te node J is ancestor te node G. 8

9 Te node wic occurs next to some oter node. In our example Node D is descendant te node H. 27. Wat is meant by Heigt-Balanced Trees? Heigt-balanced trees are also called VL trees. In a binary searc tree, te balance factor a node is te dept its left subtree minus te dept its rigt subtree. ny binary searc tree in wic te balance factor every node is -1, 0, or +1 is an VL tree. Te empty tree is an VL tree. Terefore, an VL tree as te following properties: 1. Te sub-trees every node differ in eigt by at most one. 2. Every sub-tree is an VL tree. For example, 28. For te tree Fig : a. Wic node is te root node? b. Wic node are leaves? 9

10 c. Compute te eigt. d. Compute te dept. B C D E F G H I J Fig a. Te unique node in te tree. Root node is node b. Tese are te terminal nodes te tree. Leaves are Node H, E, F, I, and J. c. Heigt te tree is 4. d. Dept te tree is 4. Because dept is equal to eigt. 29. Wat are te applications trees? pplications trees are, Trees are used in te representation sets. Decision-making. E.g.: Eigt coins problem. Computer-games suc as cess, tic-tac-toe, ceckers etc., are created and executed using trees. Manipulation aritmetic expression. Symbol table construction. Syntax-analysis. 30. Sow te result inserting 3, 2, 5, 6, 9 and 4 into an initially empty VL tree. Sow te tree after eac insertion. 10

11 31. Starting wit te following VL tree, sow te tree after deleting 20.Sow eac step. 32. Define VL Trees? n VL (delson-velskii and Landis)tree is a binary searc tree wit a balance condition. Te balance condition must be easy to maintain, and it ensures tat te dept te tree is O(logN).Te simplest idea require tat te left and rigt subtrees ave te same eigt. 33. Define B-Trees? B-tree order M is a tree wit te following structural properties: Te root is a leaf or as been 2and M cildren. ll nonleaf nodes (except te root) ave between [M/2] and M cildren. ll leaves are at te same dept. 11

12 34. Define splay tree? splay tree is a self-adjusting searc algoritm for placing and locating files (called records or keys) in a database. Te algoritm finds data by repeatedly making coices at decision points called nodes. 12

13 16 MRKS Q& 1)a)Explain ow to convert general tree into Binary tree. (pr/may2004)(or) b) Write an algoritm to convert general tree into Binary tree. Conversion general tree to Binary tree lgoritm 1. ssign Head node. LLINK=NULL RLINK=connect to itself LEVEL=0 LOC=HED. 2.Read level no, name from general tree 3.Create node structure for new i/p DT=NME LLINK=RLINK=NULL 4.Comparision b/w te previous level wit current level If (previous node level < current node level) LLINK (previous node) = current node Else a) Wile (previous node level > current node level) Move previous End wile b) [Previous node level = current node level] RLINK (previous node)= current node. 5.Repeat from step 2 for all i/p s. 6.End E.g. {Use Rigt eavy general tree} Level no name 1) 1 2 B 2 C 3 D B C 3 E 3 F E.G. D Level no E F name 2) G 1 G 2 H 2 I 2 J H I J 13

14 Explanation for above algoritm: 1.ssign Head node NULL HED. 2. New i/p, read level no, name 1, Previous node level =0 Current node level = 1 1 > 0 link previous to current node N N Head N N 3. Read level no, name 2,B N B N Previous node level =1 Current node level = 2 2 > 1 link previous to current node Head N N B N 4. Read level no, name 2,C N C N Previous node level =2 Current node level = 2 2 = 2 Rigt Link previous to current node HED N N 5. Read level no, name 3,D B N D N N C N 14

15 Previous node level =2 Current node level =3 3 > 2 Link previous to current node HED N N B C N N D N 6. Read level no, name 3,E N E N Previous node level =3 Current node level = 3 3 = 3 Rigt Link previous to current node HED N N B C N N D E N 7. Read level no, name 3,F N F N Previous node level =3 Current node level =3 3 = 3 Rigt Link previous to current node HED 15

16 N N B C N N D E F N LEFT 8. Read level no, name 1,G Previous node level =3 Current node level =1 3 >1 (find up to same level) Move previous up to it reac level 1. 1 = 1 Rigt Link previous to current node N G N HED N G N LEFT 9. Read level no, name 2,H N H N Previous node level =1 Current node level =2 2 > 1 Link previous to current node HED G N 16

17 LEFT N H N 10. Read level no, name 2,I N I N Previous node level =2 Current node level =2 2 = 2 Rigt Link previous to current node HED G N LEFT N H N I N 11. Read level no, name 2,J N J N Previous node level =2 Current node level =2 2 > 1 Rigt Link previous to current node HED G N LEFT N H N I N J N 17

18 FULL BINRY TREE HED G N N H N I N J N N B C N N D E F N 2) a)explain Binary Tree Traversal in C. (OR) b)give and explain te various tree traversal algoritms wit example. (Nov/Dec 2003). c)wat are te different tree traversals? Explain wit examples. (Nov/Dec 2005) Traversing te tree means visiting eac node exactly once. Because tere are six ways to traverse a tree. For tese tree traversals we will use some notations as follows: 1. L means move to te Cild. 2. R means move to te Rigt Cild. 3. D means te Root/Parent node. Now, wit tis L, R, D one can ave six different combinations L, R, and D nodes. Suc as LDR, LRD, DLR, DRL, RLD, RDL. But from computing point view we will ave tree different ways to traversing a tree. Tese tree combinations will be LDR, LRD, and DLR. Tose are called In order; Post order, Pre order traversal respectively. In order traversal: Print 3 rd 18

19 Print 2 nd B D Print 4 t Print 1 st C E Print tis node at last In order traversals are tracing te sequence LDR tis type traversal is called in order traversal. Te basic principal is to traverse left sub-tree ten root and ten te rigt sub-tree. C B D E is te in order traversal (ie). 1 st we go towards te left most node. (ie) C. So print tat te node C. Ten go back to te node B and print B. Ten root node ten move towards te rigt print D and finally E. lgoritm for In order traversal INORDER (Root) 1. [Start from root] ptr=root 2. [Ceck te ptr value] If (ptr! =NULL) ten i) INORDER (Ptr. Lc) /*Traverse te left sub-tree ii) VISIT (ptr) /* visit te node iii) INORDER (Ptr. Rc) /* Traverse te rigt sub-tree 3. End if 4. Stop Pre order traversal: Print 1 st Print 2 nd Print 4 t B D C E 19

20 Print 3 rd Print tis node at last B C D E is te preorder traversal te above figure. We are following DLR pat (ie) data at te root node will be printed first, ten we move on te left sub-tree and go on printing te data at te node & ten to te rigt subtree. Follow te same DLR principal at eac sub-tree & go on printing te data accordingly. lgoritm for Pre order traversal PREORDER (Root) 1. [Start from root] ptr=root 2. [Ceck te ptr value] If (ptr! =NULL) ten i) VISIT (ptr) /* visit te node ii) PREORDER (Ptr. Lc) /*Traverse te left sub-tree iii) PREORDER (Ptr. Rc) /* Traverse te rigt sub-tree 3. End if 4. Stop Post order traversal: Print tis node at last Print 3 rd B E Print 4 t Print 1 st C D Print 2 nd From te figure post order sequence is C D B E. In post order traversal we are following te LRD principle (ie) move to te left most node ceck if rigt sub-tree is tere or not ten print te left most, if rigt sub-tree is tere move towards te rigt most node. Te key idea ere is tat at eac sub-tree we are following te LRD principal and print te data according. 20

21 lgoritm for Post order traversal POSTORDER (Root) 1. [Start from root] ptr=root 2. [Ceck te ptr value] If (ptr! =NULL) ten i) POSTORDER (Ptr. Lc) /*Traverse te left sub-tree ii) POSTORDER (Ptr. Rc) /* Traverse te rigt sub-tree iii) VISIT (ptr) /* visit te node 3. End if 4. Stop 3)a)Wat are te different ways Binary Tree representations? (OR) b) How is a binary tree represented using an array? Give example. (Nov/Dec 05) Tere are two ways represent ting te Binary Tree. Sequential Representation Linked Representation. Sequential Representation Eac node is sequentially arranged from top to bottom & front left to rigt. Let us understand tis matter by numbering eac node. Te numbering will start from root node and ten remaining nodes will give ever-increasing numbers in level wise direction. Te nodes on te same level will be numbered from left to rigt. Te numbering will us sown below in figure. 21

22 B C D E F G H I J K L M N O You will get a point tat a binary tree dept n aving 2 n-1 number nodes. In te figure te tree is aving te dept 4 and total number nodes are 15. Tus remember tat is a binary tree dept n tere will be a maximum 2 n-1 nodes. nd so if we know te maximum dept te tree tan we can represent binary tree using arrays data structure. We can use array size n. Te root will be at position 1. Its left cild will be at position 2; its rigt cild will be at position 3 and so on. Te way placing te elements in te array is by applying te formula is as below. Parent (n) = floor [(n-1)/2] were n>0 wen n is equal to 0 te root node will be placed at tat location & root node is not aving any parent. (n) =(2n +1) Rigt (n) = (2n+2) Root = =index 0 cild i.e.; n =0 B will be at 2*0+1=1 st location Similarly left rigt cild wic will be C C will be at 2*0+2=2 nd location I is at te 8 t location 2n + 2 =8 2n = 8 2 =6 2n = 6 n = 6/2 = 3 22

23 Tat means parent I is at te 3 rd location and i.e. P B C D E F G H I 0 B 1 C 2 D 3 E 4 F 5 G 6 H 7 I 8 : 23

24 dvantages Disadvantages Te only advantages wit tis type representation is tat te direct access to any node can be possible & finding te parent or left, rigt cild any particular node is fast because te random access. Te major disadvantage wit tis type representation is wastages memory. In tis type representation te maximum dept te tree as to be fixed, because we ave decide te array size. Te insertion and deletion any node in te tree will be costlier as oter nodes ave to be adjusted at appropriate positions. So tat te meaning binary tree can be preserved. Linked Representation In linked list eac node will be look like tis Rigt cild Data cild In binary tree eac node will ave left cild, rigt cild and data field. Te left cild is noting but te left link wic points to some address left sub tree were as rigt cild is also a rigt link wic points to some address rigt Sub tree. nd te data field gives te information about te node. Let us see te C structure te node to a binary tree. Type def struct node { int data; struct node *left; struct node *rigt; } bin; B C D E 24

25 dd node B dd node C dd node B dd node D E Null C Null Null D Null Null E Null dvantages: Using tis array representation as tere is no wastage memory & so tere is no need to ave prior knowledge dept te tree. Insertion & deletion, wic are te most common operation, can be done witout moving te oter nodes. Disadvantages: Tis representation does not provide direct access to a node and special algoritms are required. Tis representation needs additional space in eac node for storing te left and rigt sub-trees. 4)a)Wat is te use Find-min and Find-max procedures? Explain te Find operation for binary searc trees wit example.(or) b) Write algoritm to locate an element from binary searc tree.(nov/dec05) Find Tis operation generally requires returning a pointer to te node in tree T tat as key X or NULL if tere is no suc node. Te public routine tat returns nonzero if te Find succeeded, and sets Last Find. If te find failed, zero is returned, and Last Find point to NULL. Te structure te tree makes tis simple. If T is NULL, ten we return 0. Oterwise, if te key stored at T is X, we can return T oter wise, we make a recursive call on a T, eiter left or rigt, depending on te relationsip X to te key stored in T. Find operation for binary searc trees algoritm is given below: 25

26 Find_Min and Find_Max: Template < class Etype> Tree_Node<Etype>* Binary_Searc_Tree<Etype> Find (const Etype & X, Tree-Node<Etype>* T) const { If (T==NULL) Return NULL; Else If (x<t Element) Return Find (X, T ); /* Find te element at */ Else If (x > T Element) Return Find (X, T Rigt); /* Find te element at Rigt */ Else Return T; } To perform a Find_Min, start at te root and go left, as tere is a left cild. Te stopping point is te smallest element. Te Find_Max routine is te same; expect tat brancing be to te rigt cild. Te public interface is similar to tat te Find routine. lgoritm for Find_Min for binary Searc trees Template < class Etype> Tree_Node<Etype>* Binary_Searc_Tree<Etype> Find_Min (Tree Node < Etype>* T) const { If (T==NULL) Return NULL; Else If (T ==NULL) Return T; Else Return Find_Min (T ); } Tis is so easy tat many programmers do not boter using recursion. We will code te routines bot ways by doing Find_Min recursively and Find_Max non-recursively. Notice ow we carefully andle te degenerate case o an empty tree; altoug tis is always important to do, it is recursive programs. lgoritm for Nonrecursive implementation Find_Max for binary Searc trees 26

27 Template < class Etype> Void Binary_Searc_Tree<Etype> Find_Max (Tree_Node < Etype>* T) const { If (T! =NULL) Wile (T Rigt! = NULL) T = T Rigt; Return T; } INSERT To insert X into tree T, process down te tree as you would wit a FIND procedure. If X is found, do noting; oterwise, insert X at last position on te pat traversed Insert 5 27

28 lgoritm to insert into a binary searc tree template<class type> Void Binary_searc_Tree<Etype>:: Insert (Const Etype &X, Tree-Node<Etype> * & T) { If (T==NULL) { T=new Tree_Node<Etype>(X); If (T==NULL) Error ( out space ); } Else If (X < T ->Element) Insert (X, T->); Else If(X > T ->Element) Insert (X, T->Rigt); } 28

29 5.a) ow do you insert an element into an VL tree? (May/Jun 05) (OR) b) Write a function to generate te VL tree eigt H wit fewest nodes. (May/Jun 2005) (OR) c) Explain te single rotation and double rotation in an VL tree. (OR) d) Write te routine to perform single rotation and double rotation. VL TREES: Invented by Russian scientists G.M.delson-velsky and E.M.Landis in Definition: n VL tree is a binary tree in wic te balance factor every node is eiter 0 or 1, wic is defined as difference between te eigts te nodes left &rigt sub trees. Heigt is maintained more or less in balanced tree so maintaining anoter factor called balancing factor. Heigt for eac node is calculated by Heigt left Heigt rigt. It may ave values 0, 1 for VL tree. Heigt tree differs almost by 1. If it becomes 2 ten we ave to adjust to become 1,te node aving balance factor as 2 is called pivot node, wose balance factor 2. If te node is more tan one ( 1), ten selected as pivot node at lowest level. INSERTION General algoritm for insertion 1. Insert similarly to binary searc tree 2. Ceck for balancing factor 3. If balancing factor any node = 2 i. Find pivot node ii. Do rotation at pivot node. Insertion can be done in, 1) cild pivot node SINGLE RIGHT ROTTION 2) Rigt Rigt cild pivot node SINGLE LEFT ROTTION 3) Rigt cild pivot nod LEFT RIGHT ROTTION 29

30 4) Rigt cild pivot node RIGHT LEFT ROTTION SINGLE RIGHT ROTTION SINGLE LEFT ROTTION Mirror image eac oter SINGLE RIGHT ROTTION Wen insertion in left left cild pivot node. Before insertion: S bf (S) = +1 - = 1 bf (M) = = 0 sub tree M M Rigt sub tree M Rigt sub tree S In te given above tree, te balancing factor eac and every node in te tree is in balanced condition. So, tere is no cange.if we add or insert a new node in te left left cild, ten it becomes unbalanced so you ave rotate te nodes to become balanced tree. Tis can be done troug Single rigt rotation in te following paragraps. fter insertion: S K1 bf (S)=+1+1- = 2 pivot node bf (M)=+1-=1 M K2 Rigt S +1 M Rigt sub tree M Insertion 30

31 In te above tree, if we insert node in te left left cild, it becomes unbalanced, node S balance factor is 2 and it sould be noted as pivot node and make te rigt rotation to bring as balanced tree. Te balanced tree after single rigt rotation as given below. Here te pointer rotation can be mentioned in te box below. fter rotation bf (M)=+1 - ( + 1) = 0 M +1 M Insertion Older Rigt sub tree M S bf (S)=-=0 Rigt S So te rotation can be mentioned as K2 = K1 K1 = K2 Rigt K2 Rigt = K1 SINGLE LEFT ROTTION Wen insertion is done in rigt rigt cild pivot node In te given below tree, te balancing factor eac and every node in te tree is in balanced condition. So, tere is no cange.if we add or insert a new node in te rigt rigt cild, ten it becomes unbalanced so you ave rotate te nodes to become balanced tree. Tis can be done troug Single left rotation in te following paragraps. 31

32 Before insertion M bf (M)= - ( + 1) = - 1 M Q Q bf (Q)=-=0 Rigt Q fter insertion: M K1 bf (M)= - ( + 1+1) = - 2 pivot node M bf (Q)=- (+1) = -1 K2 Q Rigt sub tree Q +1 Q Insertion In te above tree, if we insert node in te rigt rigt cild, it becomes unbalanced, node M balance factor is 2 and it sould be noted as pivot node and make te left rotation to bring as balanced tree. Te balanced tree after single left rotation as given below. Here te pointer rotation can be mentioned in te box below. 32

33 fter rotation: Q bf (Q) = +1 ( +1) = 0 bf (M) = = 0 M M Older Q Rigt Q Insertion +1 So te rotation can be mentioned as K2 = K1 Rigt K1 Rigt = K2 K2 = K1 DOUBLE ROTTION Double rotation can be classified into two types, namely LEFT RIGHT ROTTION (L R) RIGHT LEFT ROTTION(R L) Mirror image eac oter LEFT RIGHT ROTTION (L R) Wen insertion is done in left rigt cild pivot node.to understand Rigt Rotation (LR), we can consider te given tree; te balancing factor eac and every node in te tree is in balanced condition. So, tere is no cange.if we add or insert a new node in te, left rigt cild ten it becomes unbalanced so you ave rotate te nodes to become balanced tree. Te insertion may be eiter in left cild or rigt cild left only. 33

34 Before insertion: S K1 bf (S)=(+1)- (+2) = S T T K2 V bf (T) = - = 0 Rigt T bf (V)=(+1) =1 Rigt V fter insertion: S K1 bf (S)= + 1- (+3) = -2 pivot node +1 S +1 T T K2 V bf (T) = +1- = 1 Rigt T bf (V)=(+2) =2 pivot node Rigt V Insertion 34

35 Consider Rigt sown below can be rotated in single rigt rotation tis is as follows. Before insertion: bf (V) = = 2 V T T +1 bf (T)=+1- =1 Rigt T Rigt V Insertion (RIGHT) fter rigt rotation: T T Insertion +1 Older Rigt T V Rigt V 35

36 fter rigt rotation, you ave to make left rotation to get balanced tree. Tis is given as below S bf (S)= (+2) = -2 pivot node S T Insertion +1 T bf (T) =+1 - (+1) =0 Older Rigt T V bf (V) = =0 Rigt V (LEFT) T bf (T) =+1 - (+1) =0 S bf(s)=-(+1)=-1 V bf (V) = =0 S Older T Insertion +1 Older Rigt T Rigt V 36

37 RIGHT LEFT ROTTION (R L) Wen insertion is done in Rigt cild pivot nodeto understand Rigt Rotation (RL), we can consider te given tree; te balancing factor eac and every node in te tree is in balanced condition. So, tere is no cange.if we add or insert a new node in te, rigt left cild ten it becomes unbalanced so you ave rotate te nodes to become balanced tree.te insertion may be eiter in left cild or rigt cild rigt only. Before insertion: bf (S)=+2-(+1) S M bf(m)=+1-(+1) Q bf (Q)=- =0 Rigt S M Q Rigt Q fter insertion: S bf (S)=+3-(+1)=2 bf(m)=+1-(+2)=-1 M bf(q)=+1-=1 Q Rigt S M +1 Rigt Q Q Insertion 37

38 Consider sown below can be rotated in single rigt rotation tis is as follows. Before insertion: M Bf(M)=--2=-2 T +1 Rigt Q Q bf(q)=+1-=1 Rigt Q Insertion fter rigt rotation: (LEFT) Q M M Older Q Insertion +1 Rigt Q fter left rotation, you ave to make rigt rotation to get balanced tree. Tis is given as below 38

39 fter left rotation: S Q M M Older Q Insertion +1 Rigt Q Rigt Sub tree S +1 (RIGHT) Q bf (Q) =+2 - (+1) =1 M bf(m)=-(+1)=-1 S bf (S) = =0 M Older Q Insertion +1 Older Rigt Q Rigt S 39

40 6) Explain B-Tree. B-tree order m is a tree wit te following structural properties: o Te root is eiter a leaf or as between 2 and M cildren. o ll nonleaf nodes (except te root) ave between M/2 and M cildren. o ll leaves are at te same dept. ll data is stored at te leaves. Contained in eac interior node are pointers P 1, P 2,..., P M to te cildren, and values k 1, k 2,..., k m - 1, representing te smallest key found in te subtrees P 2, P3,..., P M respectively. For every node, all te keys in subtree P 1 are smaller tan te keys in subtree P 2, and so on. Te leaves contain all te actual data, wic is eiter te keys temselves or pointers to records containing te keys. Te number keys in a leaf is also between M/2 and M. Te tree in te following figure is an example a B-tree order 4. B-tree order 4. B-tree order 4 is more popularly known as a tree, and a B-tree order 3 is known as a 2-3 tree. Te following is te 2-3 tree. 2-3 tree 40

41 Te interior nodes (nonleaves) are in ellipses, wic contain te two pieces data for eac node. das line as a second piece information in an interior node indicates tat te node as only two cildren. Leaves are drawn in boxes, wic contain te keys. Te keys in te leaves are ordered. To perform a find, we start at te root and branc in one (at most) tree directions, depending on te relation te key we are looking for to te two (possibly one) values stored at te node. To perform an insert on a previously unseen key, x, we follow te pat as toug we were performing a find. Wen we get to a leaf node, we ave found te correct place to put x. Tus, to insert a node wit key 18, we can just add it to a leaf witout causing any violations te 2-3 tree properties. If we now try to insert 1 into te tree, we find tat te node were it belongs is already full. Placing our new key into tis node would give it a fourt element wic is not allowed. Tis can be solved by making two nodes two keys eac and adjusting te information in te parent. If we insert 19 into te current tree, we make two nodes two keys eac, we obtain te following tree. 41

42 Tis tree as an internal node wit four cildren, but we only allow tree per node. So split tis node into two nodes wit two cildren. Of course, tis node migt be one tree cildren itself, and tus splitting it would create a problem for its parent (wic would now ave four cildren), but we can keep on splitting nodes on te way up to te root until we eiter get to te root or find a node wit only two cildren. If we now insert an element wit key 28, we create a leaf wit four cildren, wic is split into two leaves two cildren: 42

43 Tis creates an internal node wit four cildren, wic is ten split into two cildren. Wat we ave done ere is split te root into two nodes. Wen we do tis, we ave a special case, wic we finis by creating a new root. Tis is ow (te only way) a 2-3 tree gains eigt. Wen a key is inserted, te only canges to internal nodes occur on te access pat. Tese canges can be made in time proportional to te lengt tis pat, Te dept a B-tree is at most log [M/2] N. t eac node on te pat, we perform O(log M) work to determine wic branc to take (using a binary searc), but an insert or delete could require O(M) work to fix up all te information at te node. Te worst-case running time for eac te insert and delete operations is tus O(M log M N) = O( (M / log M ) log N), but a find takes only O(log N ). 7) Explain Treaded Trees. Te linked representation any binary tree consists more null links tan actual pointers, tere are n + 1 null links and 2n total links. clever way to make use tese null links as been devised by. J. Perlis and C. Tornton. Teir idea is to replace te null links by pointers, called treads, to oter nodes in te tree. If te RCHILD(P) is normally equal to zero, we will replace it by a pointer to te node wic would be printed after P wen traversing te tree in inorder. null LCHILD link at node P is replaced by a pointer to te node wic immediately precedes node P in inorder. 43

44 Binary tree Te binary tree wit its new treads (drawn as dotted lines). Te tree T as 9 nodes and 10 null links wic ave been replaced by treads. If we traverse T in inorder te nodes will be visited in te order H D I B E F C G. For example node E as a predecessor tread wic points to B and a successor tread wic points to. In te memory representation we must be able to distinguis between treads and normal pointers. Tis is done by adding two extra one bit fields LBIT and RBIT. LBIT(P) =1 LBIT(P) = 0 RBIT(P) = 1 RBIT(P) = 0 if LCHILD(P) is a normal pointer if LCHILD(P) is a tread if RCHILD(P) is a normal pointer if RCHILD(P) is a tread Two treads ave been left dangling in LCHILD(H) and RCHILD(G). In order tat we leave no loose treads we will assume a ead node for all treaded binary trees. Ten te complete memory representation for te above tree is sown in te following figure. Te tree T is te left subtree te ead node. We assume tat an empty binary tree is represented by its ead node as 44

45 First, we observe tat for any node X in a binary tree, if RBIT(X) = 0, ten te inorder successor X is RCHILD(X) by definition treads. If RBIT(X) = 1, ten te inorder successor X is obtained by following a pat left cild links from te rigt cild X until a node wit LBIT = 0 is reaced. Te algoritm INSUC finds te inorder successor any node X in a treaded binary tree. Memory Representation Treaded Tree procedure INSUC(X) //find te inorder succcesor X in a treaded binary tree// S RCHILD(X) //if RBIT(X) = 0 we are done// if RBIT(X) = 1 ten [wile LBIT(S) = 1 do //follow left// S LCHILD(S) //until a tread// end] return (S) end INSUC If we wis to list in inorder all te nodes in a treaded binary tree, ten we can make repeated calls to te procedure INSUC. Since te tree is te left subtree te ead node and because te coice RBIT= 1 for te ead node, te inorder sequence nodes for tree T is obtained by te procedure TINORDER. procedure TINORDER (T) //traverse te treaded binary tree, T, in inorder// HED T loop T INSUC(T) if T = HED ten return print(dt(t)) 45

46 forever end TINORDER Te computing time is still O(n) for a binary tree wit n nodes. Te insertions into a treaded tree will give us a procedure for growing treaded trees. We sall study only te case inserting a node T as te rigt cild a node S. If S as an empty rigt subtree, ten te insertion is simple and diagrammed in te following figure a. If te rigt subtree S is non-empty, ten tis rigt subtree is made te rigt subtree T after insertion. Wen tis is done, T becomes te inorder predecessor a node wic as a LBIT = 0 and consequently tere is a tread wic as to be updated to point to T. Te node containing tis tread was previously te inorder successor S. Figure b illustrates te insertion for tis case. In bot cases S is te inorder predecessor T. Te details are spelled out in algoritm INSERT_RIGHT. Insertion T as a Rigt Cild S in a Treaded Binary Tree procedure INSERT_RIGHT (S,T) 46

47 //insert node T as te rigt cild S in a treaded binary tree// RCHILD(T) RCHILD(S); RBIT(T) RBIT(S) LCHILD(T) S; LBIT(T) 0 //LCHILD(T) is a tread// RCHILD(S) T; RBIT(S) 1 //attac node T to S// if RBIT (T) = 1 ten [W INSUC(T)//S ad a rigt cild// LCHILD(W) T] end INSERT_RIGHT 8) Explain Splay Trees. Splay trees are binary searc trees tat acieve our goals by being self-adjusting in a quite remarkable way: Every time we access a node te tree, weter for insertion or retrieval, we perform radical surgery on te tree, lifting te newly accessed node all te way up, so tat it becomes te root te modified tree. Oter nodes are pused out te way as necessary to make room for tis new root. Nodes tat are frequently accessed will frequently be lifted up to become te root, and tey will never drift too far from te top position. Inactive nodes, on te oter and, will slowly be pused farter and farter from te root. It is possible tat splay trees can become igly unbalanced, so tat a single access to a node te tree can be quite expensive. Over a long sequence accesses, splay trees are not at all expensive and are guaranteed to require not many more operations even tan VL trees Te analytical tool used is called amortized algoritm analysis, amortized analysis since, like insurance calculations, te few expensive cases are averaged in wit many less expensive cases to obtain excellent performance over a long sequence operations. We perform te radical surgery on splay trees by using rotations a similar form to tose used for VL trees, but now wit many rotations done for every insertion or retrieval in te tree. In fact, rotations are done all along te pat from te root to te target node tat is being accessed. Te key idea splaying is to move te target node two levels up te tree at eac step. Consider some simple terminology: Consider te pat going from te root down to te accessed node. Eac time we move left going down tis pat, zig and zag we say tat we zig, and eac time we move rigt we say tat we zag. move two steps left (going down) is ten called zig-zig, two steps rigt zag-zag, left ten rigt zig-zag, and rigt ten left zag-zig. Tese four cases are te only possibilities in moving two steps down te pat. If te lengt te pat is odd, owever, tere will be one more step needed at its end, eiter a zig (left) move, or a zag (rigt) move. Te rotations done in splaying for eac zig-zig, zig-zag, and zig moves are sown in Figure Te oter tree cases, zag-zag, zag-zig, and zag are mirror images tese. 47

48 Splay rotations Te zig-zag case in above figure is identical to tat an VL double rotation and te zig case is identical to a single rotation. Te zig-zig case, owever, is not te same as would be obtained by lifting te target node twice wit single rotations. 48

49 Zig-zig incorrectly replaced by single rotations To avoid te error sown in above Figure, always tink lifting te target two levels at a time (except wen only a single zig or zag step remains at te end). lso, note tat it is only te nodes on te pat from te target to te root wose relative positions are canged, and tat only in te ways sown by colored dased curves in previous figure. None te subtrees f te pat (sown as T1, T2, T3, and T4 in tat figure canges its sape at all, and tese subtrees are attaced to te pat in te only places tey can possibly go to maintain te searc-tree ordering all te keys. Let us work troug an example, We start wit te top left tree and splay at c. Te pat from te root to c goes troug, f, b, e, d, c. Frome to d to c is zig-zig (leftleft, as sown by te dased oval), so we perform a zig-zig rotation on tis part te tree, obtaining te second tree in te following Figure. Note tat te remainder te tree as not canged sape, and te modified subtree is ung in te same position it originally occupied. 49

50 Example splaying For te next step, te pat from f to b to c (sown inside te dased curve in te second tree) is now a zig-zag move, and te resulting zig-zag rotation yields te tird tree. Here te subtree d, e (f te pat) does not cange its sape but moves to a new position, as sown for T3 in Splay rotation figure. In tis tird tree, c is only one step from te root, on te left, so a zig rotation yields te final tree te above figure. Here, te subtree rooted at f does not cange sape but does cange position. In tis example, we ave performed bottom-up splaying, beginning at te target node and moving up te pat to te root two steps at a time. In working troug examples by and, tis is te natural metod, since, after searcing from te top down, one would expect to turn around and splay from te bottom back up to te top te tree. Hence, being done at te end te process if necessary, a single zig or zag move occurs at te top te tree. Bottom-up splaying is essentially a two-pass metod, first searcing down te tree and ten splaying te target up to te root. In fact, if te target is not found but is ten inserted, bottom-up splaying migt even be programmed wit tree passes, one to searc, one to insert, and one to splay. In writing a computer algoritm, owever, it turns out to be easier and more efficient to splay from te top down wile we are searcing for te target node. Wen we find te target, it is immediately moved to te root te tree, or, if te searc is unsuccessful, a new root is created tat olds te target. In top-down splaying, a single zig or zag move occurs at te bottom te splaying process. 50