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CSE 326 Lecture 7: More on Search Trees

! Todays Topics:" Lazy Operations" Run Time Analysis of Binary Search Tree Operations" Balanced Search Trees

# AVL Trees and Rotations

! Covered in Chapter 4 of the text

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From Last Time: Remove (Delete) Operation! Removing a node containing X:

1. Find the node containing X2. Replace it with:

If it has no children, with NULLIf it has 1 child, with that childIf it has 2 children, with the node withthe smallest value in its right subtree,(or largest value in left subtree)

3. Recursively remove node used in 2 and 3

! Worst case : Recursion propagates allthe way to a leaf node time isO(depth of tree)

10

5 24

94

97

11

17

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Laziness in Data Structures

! A lazy operation is one that puts off work as much aspossible in the hope that a future operation will make thecurrent operation unnecessary

D a t a S t r u c t - u r e s

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Lazy Deletion! Idea: Mark node as deleted ; no need to reorganize tree

" Skip marked nodes during Find or Insert" Reorganize tree only when number of marked nodes

exceeds a percentage of real nodes (e.g. 50%)" Constant time penalty only due to marked nodes depth

increases only by a constant amount if 50% are markedundeleted nodes (N nodes max N/2 marked)

! Modify Insert to make use of marked nodes whenever

possible e.g. when deleted value is re-inserted! Can also use lazy deletion for Lists

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Run Time Analysis of BST operations

! All BST operations (except MakeEmpty) are O(d), where d isthe depth of the accessed node in the tree" MakeEmpty takes O(N) for a tree with N nodes frees all nodes

! We know: log N ! d ! N-1 for a binary tree with N nodes" What is the best case tree? What is the worst case tree?

! Best Case Running Time of Insert/Remove/etc. = ?

! Worst Case Running Time = ?

! Average Case Running Time = ?

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The best, the worst, and the average! For a binary tree with N nodes, depth d of any node satisfies:

log N ! d ! N-1

! So, best case running time of BST operations is O(log N)

! Worst case running time is O(N)

! Average case running time = O(average value of d) = O(log N)" Can prove that average depth over all nodes = O(log N) if all

insertion sequences equally likely." See Chap. 4 in textbook for proof

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Can we do better?

! Worst case running time of BST operations is O(N)

! E.g. What happens when you Insert elements in ascending(or descending) order ?" Insert 2, 4, 6, 8, 10, 12 into an empty BST

! Problem : Lack of balance Tree becomes highly asymmetric

! Idea: Can we restore balance by re-arranging tree according todepths of left and right subtrees?"

Goal: Get depth down from O(N) to O(log N)

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Idea #1: Achieving the perfect balance! First try at balancing trees: Perfect balance

" Re-arrange to get a complete tree afterevery operation

! Recall: A tree is complete if there are noholes when scanning from top tobottom, left to right

! Problem : Too expensive to re-arrange" E.g. Insert 2 in the example shown

! Need a looser constraint

6

4 9

1 5 8

5

2 8

1 4 6 9

Insert 2 &make treecomplete

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Idea #2: Leave it to the professionals

! Many efficient algorithms exist for balancing trees inorder to achieve faster running times for the BSToperations" Adelson-Velskii and Landis (AVL) trees (1962)" Splay trees and other self-adjusting trees (1978)" B-trees and other multiway search trees (1972)

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AVL Trees! AVL trees are height-balanced binary

search trees

! Balance factor of a node = height(leftsubtree) - height(right subtree)

! An AVL tree can only have balancefactors of 1, 0, or 1 at every node" For every node, heights of left and

right subtree differ by no more than 1

" Height of an empty subtree = -1

! Implementation : Store currentheights in each node

6

4

1

8

7

21

1

9

6

4

1

8

7

-1

1-(-1) = 21

(-1)-0 = -1

9

Heights

Balancefactors

0

0

0

0

3

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The good news about AVL Trees

! Can prove: Height of an AVL tree of Nnodes is always O(log N)

! How? Can show:" Height h ! 1.44 log(N+2)-0.328" Prove using recurrence relation for

minimum number of nodes S(h) in anAVL tree of height h:S(h) = S(h-1) + S(h-2) + 1

" Use Fibonacci numbers to get bound

on S(h) bound on height h" See textbook for details

6

4 9

1 5 8

0

10

00

0

0

1 -1

0

6

4

17

90Height =O(log N)

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The really good news about AVL Trees! Can prove: Height of an AVL tree of N

nodes is always O(log N)

! All operations (e.g. Find, Removeusing lazy deletion, etc.) on an AVLtree are O(log N)

! except Insert" Why is Insert different? 0

1 -1

0

6

5

4

7

90

Insert 3

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The bad news about AVL Trees

0

1 -1

0

6

5

4

7

90

Insert 31

2 -1

1

6

5

4

7

90

0 3

No longer an AVL tree(i.e. not balanced anymore)

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Restoring Balance in (the life of) an AVL Tree! Problem : Insert may cause balance factor to become 2 or 2

for some node on the path from insertion point to root node

! Idea : After Inserting the new node,1. Back up to root updating heights along the access path2. If Balance Factor = 2 or 2, adjust tree by rotation

around deepest such node.1

2 -1

1

6

5

4

7

90

0 3

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Rotating to restore Balance: A Simple Example

0

1 -1

0

6

5

4

7

90

Insert 3 1

2 -1

1

6

5

4

7

90

0 3

0

0 -1

0

6

4

3

7

9 05 0

Rotate

AVL Not AVL AVL

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BF = 1

BF = 0

BF = 0 BF = 0

BF = 1

BF = 0

BF = 0BF = 0

BF = 0

Tree before insertion(BF = Balance Factor)

Various Cases of Insertion

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BF = ?

BF = ?

BF = 0BF = ?

BF = 0

Tree after insertion

BF = 0

Outside Case

BF = ?

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BF =

BF =

BF = BF =

BF =

BF =

BF =BF =

BF =

Tree after insertionBF = 0

Inside Case

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Let the node that needs rebalancing be " .

There are 4 cases:Outside Cases (require single rotation ) :

1. Insertion into left subtree of left child of " .2. Insertion into right subtree of right child of " .

Inside Cases (require double rotation ) :3. Insertion into right subtree of left child of " .4. Insertion into left subtree of right child of " .

Rebalancing is performed through four separate rotationalgorithms .

Insertions in AVL Trees

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j

k

X Y

Z

Consider a validAVL subtree

Insertions in AVL Trees: Outside Case

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j

k

X Y

Z

Inserting into Xdestroys the AVLproperty


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j

k

X Y

Z

Do a right rotation


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j

k

X Y

Z

Do a right rotation


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jk

X Y Z

Right rotation done!


AVL property has been restored!(Left rotation is mirror symmetric)

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j

k

X Y

Z

Insertions in AVL Trees: Inside Case

Consider a validAVL subtree

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Inserting into Ydestroys theAVL property

j

k

X Y

Z


Does right rotationrestore balance?

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jk

X

Y Z

Right rotationdont do nothinto restore balance


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Consider the structureof subtree Y j

k

X Y

Z


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j

k

X V

Z

W

i

Y = node i andsubtrees V and W


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j

k

X V

Z

W

i


Lets try a left-rightdouble rotation . . .

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j

k

X V

Z W

i

Steps for Left-RightDouble Rotation


1. Left Rotation :Adjust relevantpointers

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2. Right Rotation :Adjust relevantpointers

3. Make i the root jk

X V Z W

i

Balance has been restored!

(Right-left case is mirror-symmetric)

Insertions in AVL Trees: Inside CaseSteps for Left-RightDouble Rotation

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AVL Tree Exercise

! Insert 8, 1, 0, 2 in that order into following AVL tree:

4

3 6

5 7

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Arguments for AVL trees:

1. Search is O(log N) since AVL trees are always balanced .2. The height balancing adds no more than a constant factor to

the speed of insertion.

Arguments against using AVL trees :

1. Difficult to program & debug; more space for height info.2. Asymptotically faster but can be slow in practice.3. Most large searches are done in database systems on disk and

use other structures (e.g. B-trees ).4. May be OK to have O(N) for a single operation if total run

time for many consecutive operations is fast (e.g. Splay trees).

Pros and Cons of AVL Trees

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Next Class:

All about Splaying

To Do:

Finish Reading Chapter 4

raolect7

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