Binary Search Tree - Best Time. AVL Trees. Binary Search Tree - Worst Time. Balanced and unbalanced BST

Transcription

1 AL Trees CSE Data Structures Unit Reading: Section 4.4 Binary Searc Tree - Best Time All BST operations are O(d), were d is tree dept minimum d is d = log for a binary tree N wit N nodes at is te best case tree? at is te worst case tree? So, best case running time of BST operations is O(log N) Binary Searc Tree - orst Time orst case running time is O(N) at appens wen you Insert elements in ascending order? Insert:, 4,, 8,, into an empty BST Problem: Lac of balance : compare depts of left and rigt subtree Unbalanced degenerate tree Balanced and unbalanced BST Is tis balanced? 7 7 4

2 Approaces to balancing trees Don't balance May end up wit some nodes very deep Strict balance Te tree must always be balanced perfectly Pretty good balance Only allow a little out of balance Adust on access Self-adusting Balancing Binary Searc Trees Many algoritms exist for eeping binary searc trees balanced Adelson-elsii and Landis (AL) trees (eigt-balanced trees) Splay trees and oter self-adusting trees B-trees and oter multiway searc trees Perfect Balance ant a complete tree after every operation tree is full except possibly in te lower rigt Tis is expensive For example, insert in te tree on te left and ten rebuild as a complete tree 8 Insert & complete tree AL - Good but not Perfect Balance AL trees are eigt-balanced binary searc trees Balance factor of a node eigt(left subtree) - eigt(rigt subtree) An AL tree as balance factor calculated at every node For every node, eigts of left and rigt subtree can differ by no more tan : For every node t (t.left)-(t.rigt) {-,, } Store current eigts in eac node 8

3 Heigt of an AL Tree N() = minimum number of nodes in an AL tree of eigt. Basis N() =, N() = Induction N() = N(-) + N(-) + Solution (recall Fibonacci analysis) N() > φ (φ.) - - Heigt of an AL Tree N() > φ (φ.) Suppose we ave n nodes in an AL tree of eigt. n > N() (because N() was te minimum) n > φ ence log φ n > (relatively well balanced tree!!) <.44 log n (i.e., Find taes O(logn)) 9 Tree A (AL) Node Heigts Tree B (AL) 8 eigt of node = balance factor = left - rigt empty eigt = - Node Heigts after Insert 7 Tree B (AL) 8 Tree C (not AL) 8 eigt of node = balance factor = left - rigt empty eigt = - 7 balance factor -(-) = -

4 Insert and Rotation in AL Trees Insert operation may cause balance factor to become or for some node only nodes on te pat from insertion point to root node ave possibly canged in eigt So after te Insert, go bac up to te root node by node, updating eigts If a new balance factor (te difference left - rigt ) is or, adust tree by rotation around te node Single Rotation in an AL Tree Insertions in AL Trees Let te node tat needs rebalancing be α. Tere are 4 cases: Outside Cases (require single rotation) :. Insertion into left subtree of left cild of α.. Insertion into rigt subtree of rigt cild of α. Inside Cases (require double rotation) :. Insertion into rigt subtree of left cild of α. 4. Insertion into left subtree of rigt cild of α. Te rebalancing is performed troug four separate rotation algoritms. AL Insertion: Outside Case Consider a valid AL subtree + +

5 AL Insertion: Outside Case Becomes + + Inserting into destroys te AL property at node (+) - AL Insertion: Outside Case + Do a rigt rotation 7 8 Single rigt rotation + Do a rigt rotation 9 Outside Case Completed + Rigt rotation done! ( Left rotation is mirror symmetric) AL property as been restored!

6 AL Insertion: Inside Case Consider a valid AL subtree + + AL Insertion: Inside Case Inserting into destroys te AL property at node Becomes + + Does rigt rotation restore balance? AL Insertion: Inside Case + Rigt rotation does not restore balance now is out of balance AL Insertion: Inside Case Consider te structure of subtree + 4

7 AL Insertion: Inside Case = node i and subtrees and + i or - + AL Insertion: Inside Case i e will do a left-rigt double rotation... Double rotation : first rotation left rotation complete Double rotation : second rotation Now do a rigt rotation i i 7 8

8 Double rotation : second rotation + + or - i double rotation complete Balance as been restored + 9 Implementation left igt ey rigt Anoter possible implementation: do not eep te eigt; ust te difference in eigt, i.e. te balance factor (,,-). In bot implementation, tis as to be modified on te pat of insertion even if you don t perform rotations Once you ave performed a rotation (single or double) you won t need to go bac up te tree Single Rotation RotateFromRigt(n : reference node pointer) { p : node pointer; p := n.rigt; n n.rigt := p.left; p.left := n; p n := p } Double Rotation Implement Double Rotation in two lines. DoubleRotateFromRigt(n : reference node pointer) {???? n } e also need to modify te eigts or balance factors of n and p Insert

9 Insertion in AL Trees Insert at te leaf (as for all BST) only nodes on te pat from insertion point to root node ave possibly canged in eigt So after te Insert, go bac up to te root node by node, updating eigts If a new balance factor (te difference left - rigt ) is or, adust tree by rotation around te node Insert in BST Insert(T : reference tree pointer, x : element) : integer { if T = null ten T := new tree; T.data := x; return ;//te lins to //cildren are null case T.data = x : return ; //Duplicate do noting T.data > x : return Insert(T.left, x); T.data < x : return Insert(T.rigt, x); endcase } 4 Insert in AL trees Example of Insertions in an AL Tree Insert(T : reference tree pointer, x : element) : integer{ int temp; if T = null ten T := new tree; T.data := x; T.eigt=; return ; else { case T.data = x : return ; //Duplicate do noting T.data > x : temp = Insert(T.left, x); if ((eigt(t.left)- eigt(t.rigt)) = ){ if (T.left.data > x ) ten //outside case T = RotatefromLeft (T); else //inside case T = DoubleRotatefromLeft (T); return temp; } T.data < x : temp = Insert(T.rigt, x); code similar to te left case Endcase } T.eigt := max(eigt(t.left),eigt(t.rigt)) +; return ; } Insert, 4

10 Example of Insertions in an AL Tree Single rotation (outside case) 4 Now Insert 4 Imbalance Now Insert Double rotation (inside case) AL Tree Deletion Imbalance Insertion of Similar but more complex tan insertion Rotations and double rotations needed to rebalance Imbalance may propagate upward so tat many rotations may be needed. 9 4

11 Pros and Cons of AL Trees Arguments for AL trees:. Searc is O(log N) since AL trees are always balanced.. Insertion and deletions are also O(logn). Te eigt balancing adds no more tan a constant factor to te speed of insertion. Double Rotation Solution DoubleRotateFromRigt(n : reference node pointer) { RotateFromLeft(n.rigt); n RotateFromRigt(n); } Arguments against using AL trees:. Difficult to program & debug; more space for balance factor.. Asymptotically faster but rebalancing costs time.. Most large searces are done in database systems on dis and use oter structures (e.g. B-trees). 4. May be OK to ave O(N) for a single operation if total run time for many consecutive operations is fast (e.g. Splay trees). 4 4