AVL Trees

AVL comes from the authors

AdeIVelki and Landis

ReeaA BST is

- a binary tree- with nudes labelled by keys- for every two nodes v.v :

. if he is in the left subtree of v

then key (a) a key 4)- if u is in the right subtree of 4

then key Cu) > key4)

An AVL Tree is A BST,that satisfies

the AVL Tree height invariant :

for every node v ,

/ height(left - height(right d) IE1

(We define height (left to be -1 if

left G) does not exist;sin . for height( rightCut) )

Operations : BST operations , modified tomaintain the balance invariant .

Unbalanced suzb-trees are " repaired

"

using BST

rotation-

right rotation

⇒ a

④ ¥④⇐ ④⑥ ¥left rotation

at 3

RotationspreserreBSTOrderInvariaut.vewant to use BST rotations to "

repair" the Art balance

- We must be sure rotations preserve the BST invariants .

Obviously , after a rotation we still have a binarytree

labelled with keys . To see the order invariant is

preserved consider : rightrotate ④at

7¥ 170i 'E¥

Suppose ux is a key in TI , y in T2 and 2- inT3

.

According to the B5T order invariant , in bothtrees we must have:Xa M L y c u a Z .

AVLTreeinsertionl.baa BST insertion .

2 .

Let v be the unbalanced nodeof greatest

depth (if there is one ) .

3. Repair the balancecondition at v."

Consider

"

4 cases for "repair"

:

"

ousted Vunbatauud" inside

"

-* V o

÷±÷÷: :*to

= ¥74 Y*

=D1 5IY-6

To fix the "outside " case : 1 rotation at V :

.

aes¥fixthe

To fix the " inside , we use tworotatiou¥

to b

Insertion here ! I hereP

7

To fix the " inside , we use,

tworotatioyi.nu

to b

III.In!*. It right:*

( like outside case)

To fix the " inside , we use tworotas :

to b

¥¥a¥¥*. It right:*.⇒i÷.

To fix the " inside , we use tworotas :

*to to

II:!⇒. It right:*

÷:::÷÷÷÷i: 'right4)← aleft (a)←T3

right (b)←T2

E

Amaximallyunbalancedheight-5AVLTree-N.SI¥35µ **.

lomplexityofAVTTreeoperati.NLWe want a bound on the height of an n- node AVLTree .

Let Nlh) = min . # of nodes in an AVL tree of height h .Observe : N(o) = I ←htt

N( 1) = 2 In→To← h or h- I

Nlh) = It N (h - 1) t N (h - 2)

÷:*, D !!> 23N ( h - 6)%

52in ( h - 2 i)0

5£" Nlo) ( since h -hi ⇒ i. %)

= 2h12

Claim : N Ch) > 2h12,for all h 7 I .

H12

ProotthatNlh)z2_By ind. on h .

Basis : NC0) =L 720=1 ;N4)= 2=2

"--Ma

II. Suppose that , for some H71 , Nch) >2h"

.

II. Show that N(htt) > zlht'k:

NlhtDz2Nlh-D ←B""

> 2.2" 7µmIt §4=2

shorter taller subtree+1+2

subtree of of height h= 2T height

(htt)th- I or h

= 2 ✓-

lomplaxityofAVLTreesearchkinser.ttor any

AVL tree with n nodes and height is > 1,

we have: N1D ⇒ 2%so log Nln) > %or in I 2

"

log.NL#a2logznso h = Of log n)

⇒ search takes time 0( log n)

Insert takes time 0 ( login) .

Aside : Base of logs in 0L ).

-

Why : hi =0( log n) not h=0( login) ?

Change of base :n

Iogbn =t%a-

= (%) Logan↳gab e constant

Change of base is multiplication by a constant,s.

"

disappears"

inside a big- 0h .

fH=0( login) ⇒ fN=0( login) for any K7 0KH .

(corrected)

Comparison of n k 26gin-• Worst-case height of BST vs AVL tree .

n Rogen÷100 13

1000 20

104 27106 40108 53

10"

66

1,000,000,000,000 = 10k 80

( 10K bytes a Terabyte , so this is big but not crazy )

AVLTreeRemov1) Do BST removal .

Cage1 : key k is at a leaf L i delete L.

Cage 2 " K is at a node V that has 1 child L.

- the child L must be a leaf .- move the key on L to v

and delete L.

'⑤Esi←target 6 =y¥¥→ '①

Cases : la is at a node v with 2 children , and successorL.

• more key at successor L of K to V• L has eithe 0 or I child : delete itas in case 1 a case2.The removal may have caused some

nodes to become

unbalelanced .

• Notice that,in each case, we deleted a leaf. We use

this fact in describing how to re -balance .

AVLTreeRemoral.IS.

w ← parent of deleted leaff-ofeach node a = w . . -

root on path from W . - - root){. if u is unbalanced

- Let T be the subtree rooted at m

- rebalance u using suitable rotations

g.if height of T did not get smaller, exit .

*

Correctness of the algorithm involves two properties :a) There is at most 1 unbalanced model after deletion

b) Rebalancing may make an ancestor unbalanced .

Removal of an AVL Tree leaf causes at most 1

node to become unbalanced .

^\-

I:÷:÷:÷::" (÷¥. :* ¥E:÷:3unbalanced node nearest to •.

node #

ANAhktreeremoralthatikustrat.es:1. Need to re - balance after removal2. Re-balancing may reduce height of a sub -tree

remove④

¥¥ ④I⑤ ② ⑤

/ µ rightat Io

Io I

5M¥ ④⑤ ⑤ '⑤

Amtrak)

End.