AVL trees. AVL trees
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2 Dnamic set DT dnamic set DT is a structure tat stores a set of elements. Eac element as a (unique) ke and satellite data. Te structure supports te following operations. Searc(S, k) Return te element wose ke is k. Insert(S, ) dd to S. Delete(S, ) Remove from S (te operation receives a pointer to ). Minimum(S) Return te element in S wit smallest ke. Maimum(S) Return te element in S wit largest ke. Successor(S, ) Return te element in S wit smallest ke tat is larger tan.ke. Predecessor(S, ) Return te element in S wit largest ke tat is smaller tan.ke.
3 Motivation In a binar searc tree, all operation take Θ() time in te worst case, were is te eigt of te tree. Te optimal eigt of a binar searc tree is log n. Even if we start wit a balanced tree, insertion/deletion operations can make te tree unbalanced. n VL tree is a special kind of a binar searc tree, wic is alwas kept banalced.
4 VL tree n VL tree is a binar searc tree suc tat for ever node, eigt(.left) eigt(.rigt) 1 (we assume tat eigt(null) = 1) are named after teir inventors, Georg delson-velsk and Evgenii Landis.
5 Te eigt of an VL tree Teorem Te eigt of an VL tree is Θ(log n). Let n k be te minimum number of nodes in an VL tree wit eigt k. n = 1.
6 Te eigt of an VL tree Teorem Te eigt of an VL tree is Θ(log n). Let n k be te minimum number of nodes in an VL tree wit eigt k. n = 1. n 1 =.
7 Te eigt of an VL tree Teorem Te eigt of an VL tree is Θ(log n). Let n k be te minimum number of nodes in an VL tree wit eigt k. n = 1. n 1 =. n = = 4. 1
8 Te eigt of an VL tree Teorem Te eigt of an VL tree is Θ(log n). Let n k be te minimum number of nodes in an VL tree wit eigt k. n = 1. n 1 =. n = = 4. n = = 7. n k = n k 1 + n k
9 Te eigt of an VL tree n k = F k+ 1 were F k is te k-t Fibonacci number. Te proof is b induction: ase: n = 1 = F 1, n 1 = = F 4 1. n k = n k 1 +n k +1 = (F k+ 1)+(F k+1 1)+1 = F k+ 1 It is known tat ( F k = 1+ 5 ) k ( 5 1 ) k 5 ( 1+ 5 ) k 1 5 Terefore, ( 1+ ) k+ ( ) k+ 5 n k ( ) k log (1+ 5)/ 5(nk ) log n k. Te eigt of an VL tree wit n nodes is log n.
10 Implementation node in an VL tree as te same fields defined for binar searc tree (ke, left, rigt, p). dditionall, ever node as a field eigt wic stores te eigt of te node.
11 Searcing Te Searc operation is andled eactl like in regular binar searc trees. Time compleit: Θ() = Θ(log n).
12 Insertion/Deletion Insertion and deletion are done b first appling te insertion/deletion algoritm of binar searc trees. fter te insertion/deletion, te tree ma not be balanced, so we need to correct it. Te time compleit of insertion/deletion in VL tree is Θ(log n).
13 Insertion fter a new leaf is inserted, te eigt of some of its ancestors increase b 1. Te eigts of te oter nodes are uncanged
14 Insertion Te eigt fields of te nodes can be updated b going up te tree from te inserted leaf, and for eac ancestor v of te leaf perform v.eigt = 1 + ma(v.left.eigt, v.rigt.eigt)
15 Insertion Some of te ancestors of te leaf ma become unbalanced
16 Let be te lowest node on te pat from te new leaf to te root wic is unbalanced (if doesn t eist we are done). Tere are 4 cases. In Case 1 suppose tat te new leaf is in te subtree of.left.left. Let =.left. Let be te eigt of te rigt cild of. Since was balanced before te insertion, te eigt of before te insertion was 1// //+1 C New leaf
17 Te insertion eiter increases te eigt of a node or doesn t cange te eigt. Since is now unbalanced, te onl possible case is tat te eigt of is + 1 before te insertion, and + afterward. Te eigt of before te insertion is +, and + afterward C New leaf
18 Te insertion eiter increases te eigt of a node or doesn t cange te eigt. Since is now unbalanced, te onl possible case is tat te eigt of is + 1 before te insertion, and + afterward. Te eigt of before te insertion is +, and + afterward C New leaf
19 fter te insertion, as a cild wit eigt + 1. Tis cild must be te left cild. Te eigt of before te insertion is. Te eigt of te rigt cild of is C +1 New leaf
20 To fi te imbalance of, perform te following operation called rigt rotation. Te rotation operation doesn t cange te inorder of te nodes. Terefore, te new tree is a valid binar searc tree C C
21 Te eigts of and after te rotations are + 1 and +. fter te rotation, and are balanced C C
22 ssume ad a parent te before te rotation, and denote it b u. Let w be te sibling of before te rotation. Te eigt of w is + 1/ + / +. If te eigt of w is + 1, ten u is unbalanced after te insertion (before te rotation). w +1/+/ u +4 +/+/ C w u +1/+/+ +1 +/+/ C
23 fter te rotation, te eigt te sibling of w (node ) is +, wic is equal to te eigt of te sibling of w before te rotation. Terefore, u is balanced. Te eigt of u after te rotation is te same as te eigt before te insertion. Repeating tese arguments, ever ancestor of u is balanced and as same eigt as before te insertion. w +1/+/ u +4 +/+/ C w u +1/+/+ +1 +/+/ C
24 Eample In tis eample, =.
25 In Case, suppose tat te new leaf is in te subtree of.left.rigt. Let =.left. Performing a rotation on and does not work C C
26 Let z =.rigt. Perform a double rotation on,, z. Te double rotation doesn t cange te inorder of te nodes. fter te double rotation,,, z are balanced. Moreover, te eigt of z is te same as te eigt of before te insertion, and terefore all ancestors of z are balanced. + + z z D C D C -1-1
27 Case is wen te new leaf is in te subtree of.rigt.rigt, and Case 4 is wen te new leaf is in te subtree of.rigt.left. Case and Case 4 are smmetric to Case 1 and Case. Case is sown below C +1 C +1
28 Deletion fter a node is deleted, te eigts of some of its ancestors decrease b 1. Te eigts of te oter nodes are uncanged. single ancestor of te deleted node can become unbalanced
29 Let be te unbalanced node (if doesn t eist we are done). ssume tat te deleted node is in te subtree of.rigt. Let =.left. Since was balanced before te deletion, te eigt of.rigt before te deletion was 1// //+1 C
30 Te deletion eiter decreases te eigt of a node or doesn t cange te eigt. Since is now unbalanced, te onl possible case is tat te eigt of.rigt is 1 before te insertion, and afterward. Te eigt of is + 1 (te insertion doesn t cange te eigt). C - -1
31 Te deletion eiter decreases te eigt of a node or doesn t cange te eigt. Since is now unbalanced, te onl possible case is tat te eigt of.rigt is 1 before te insertion, and afterward. Te eigt of is + 1 (te insertion doesn t cange te eigt). +1 C - -1
32 Since as eigt and it is balanced, one of s cildren as eigt 1 and te oter cild as eigt 1/. In Case 1, assume tat te eigt of.left is 1, and te eigt of.rigt is 1/. +1 C /-
33 In Case 1 we perform a rigt rotation on and. fter te rotation, te eigt of is / 1 and te eigt of is + 1/. fter te rotation, and are balanced / C / /- -1/- C -
34 ssume ad a parent te before te rotation, and denote it b u. Let w be te sibling of before te rotation. Te eigt of w is / + 1/ +. u is balanced after te insertion (before te rotation). u +/+/+ u w /+1/+ +1 C - -1 w /+1/ / / /- -1/- C -
35 fter te rotation, u can become unbalanced. Tis occurs wen te eigt of w is +, and te eigt of after te rotation is. If te eigt of w is, and te eigt after te rotation is, ten te eigt of u is + and + 1 afterwards. Tis can cause an imbalance in an ancestor of u. In te two cases above, additional rotation is needed. u +/+/+ u w /+1/+ +1 C - -1 w /+1/ / / /- -1/- C -
36 In Case, assume tat te eigt of.left is, and te eigt of.rigt is 1. Let z =.rigt. In tis case we perform a double rotation of,, z. +1 z - z D /- -/- C D - -/- C -/-
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