Deletion The Two Child Case 10 Delete(5) Deletion The Two Child Case. Balanced BST. Finally
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1 Deletion Te Two Cild Cse Delete() Deletion Te Two Cild Cse Ide: Replce te deleted node wit vlue gurnteed to e etween te two cild sutrees! Options: succ from rigt sutree: findmin(t.rigt) pred from left sutree : findmx(t.left) Wt cn we replce wit? // Now delete te originl node contining succ or pred Lef or one cild cse esy! // replces Finlly Blnced BST Oservtion BST: te sllower te etter! For BST wit n nodes Averge eigt is O(log n) Worst cse eigt is O(n) Simple cses suc s insert(,,,..., n) led to te worst cse scenrio Originl node contining gets deleted // Solution: Require Blnce Condition tt. ensures dept is O(log n) strong enoug!. is esy to mintin not too strong! // Potentil Blnce Conditions. Left nd rigt sutrees of te root ve equl numer of nodes Potentil Blnce Conditions. Left nd rigt sutrees of every node ve equl numer of nodes. Left nd rigt sutrees of te root ve equl eigt. Left nd rigt sutrees of every node ve equl eigt // //
2 Te AVL Blnce Condition Left nd rigt sutrees of every node ve equl eigts differing y t most Define: lnce(x) = eigt(x.left) eigt(x.rigt) AVL property: lnce(x), for every node x Ensures smll dept Will prove tis y sowing tt n AVL tree of eigt must ve lot of (i.e. O( )) nodes Esy to mintin Using single nd doule rottions // Te AVL Tree Dt Structure Structurl properties. Binry tree property. Blnce property: lnce of every node is etween - nd Result: Worst cse dept is O(log n) Ordering property Sme s for BST // Proving Sllowness Bound Let S() e te min # of nodes in n AVL tree of eigt Clim: S() = S(-) + S(-) + Solution of recurrence: S() = O( ) (like Fioncci numers) AVL tree of eigt = wit te min # of nodes // // Testing te Blnce Property An AVL Tree We need to e le to:... dt eigt cildren NULLs ve eigt - // //
3 AVL find: AVL trees: find, insert sme s BST find. AVL insert: sme s BST insert, except my need to fix te AVL tree fter inserting new vlue. AVL tree insert Let x e te node were n imlnce occurs. Four cses to consider. Te insertion is in te. left sutree of te left cild of x.. rigt sutree of te left cild of x.. left sutree of te rigt cild of x.. rigt sutree of te rigt cild of x. Ide: Cses & re solved y single rottion. Cses & re solved y doule rottion. // // Insert() Insert() Insert() Bd Cse # Fix: Apply Single Rottion AVL Property violted t tis node (x) // Single Rottion:. Rotte etween x nd cild // Single rottion in generl Z X Y - X < < Y < < Z + X Y Z Single rottion exmple // Heigt of tree efore? Heigt of tree fter? Effect on Ancestors? //
4 Insert() Insert() Insert() Bd Cse # Fix: Apply Doule Rottion AVL Property violted t tis node (x) // Doule Rottion. Rotte etween x s cild nd grndcild. Rotte etween x nd x s new cild // Doule rottion in generl c Z W - - Y X W < <X < c < Y < < Z c - Y W X Z // Heigt of tree efore? Heigt of tree fter? Effect on Ancestors? Doule rottion, step // Doule rottion, step Imlnce t node X Single Rottion. Rotte etween x nd cild Doule Rottion. Rotte etween x s cild nd grndcild. Rotte etween x nd x s new cild // //
5 Insert into n AVL tree: e c d Single nd Doule Rottions: Inserting wt integer vlues would cuse te tree to need :. single rottion?. doule rottion?. no rottion? // Student Activity Circle your finl nswer // Student Activity Insertion into AVL tree Esy Insert. Find spot for new key. Hng new node tere wit tis key. Serc ck up te pt for imlnce. If tere is n imlnce: cse #: Perform single rottion nd exit cse #: Perform doule rottion nd exit Unlnced? Bot rottions keep te sutree eigt uncnged. Hence only one rottion is sufficient! // // Insert() Hrd Insert (Bd Cse #) Insert() Unlnced? How to fix? Single Rottion // //
6 Hrd Insert (Bd Cse #) Insert() Unlnced? How to fix? Single Rottion (oops!) // // Doule Rottion (Step #) Doule Rottion (Step #) // //
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