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### Title: Implementierung von Queues: Author: Peter Hertling Last modified by: DCC Created Date: 11/20/2003 6:07:15 AM Document presentation format: Presentaci n en ... – PowerPoint PPT presentation

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Title: Chapter 6:

1
• Chapter 6  Searching trees and more Sorting
Algorithms
• 6.1   Binnary Tree
• The Bin Tree class with traversing methods
• 6.2   Searching Trees
• 6.2.1   AVL Trees
• 6.3   HeapSort and BucketSort
• 6.3.1   HeapSort
• 6.3.2   BucketSort

2
6.1    Binary trees (Continuation)
• Characteristics
• Maximal height of a binary tree with n nodes is
n-1.
• (this is when each internal node has exactly
one child, this is in fact a linear linked list.)
• Minimal number of nodes in a binary tree of
height h is h1.
• (Ditto)

3
• Maximal number of nodes in a binary tree having
higth h N(h) 2h1 - 1  Knoten.
• proof by Induction.
• Minimal higth of a binary tree having n b nodes
• O(log n) more precisely log2 (n1)
- 1.
• Justification be h the minimal height of a
binary tree with n nodes. then   2 h - 1
N(h -1) lt n ? N(h) 2 h1 - 1
• thus
• 2 h lt n 1 ? 2 h 1
• thus
• h lt log2 (n1) ? h1

4
• Theorem In a nonempty binary tree T, whose
internal nodes have each exactly two children,
the following holds (Leaves(T))
(internal nodes(T)) 1.
• Proof by Induction over the size of the tree T.
• Base case Be T a tree wit one node. Then
• (leaves(T)) 1, (internal nodes(T)) 0. This
BC. OK.
• Induction Be T a tree with more than a node.
Then the root is an internal node. Let T1 and T2
be the right and left sub-trees. According to the
induction assumption (leaves(Ti)) (internal
nodes(Ti)) 1 for i1,2. we obtain
• (leaves(T)) (leaves(T1)) (leaves(T2))
• (internal nodes(T1)) 1 (internal
nodes(T2)) 1
• (internal nodes(T)) 1.

5
• algebra BinTree
• sorts BinTree, El, boolean
• ops
• emptyTree ? BinTree
• isEmpty BinTree ? boolean
• isLeaf BinTree ? boolean
• makeTree ? BinTree x El x BinTree ? BinTree
• rootEl BinTree ? El
• leftTree, rightTree BinTree ? BinTree
• sets
• BinTree ltgt ltL,x,Rgt L,R BinTree, x El
• functions
• emptyTree() ltgt
• makeTree(L,x,R) ltL,x,Rgt
• rootEl(lt_,x,_gt) x
• ...
• end BinTree.

6
Traversing methods for binary trees
• ops
• inOrder, preOrder, postOrder BinTree ? List
• functions
• inOrder(ltgt) ltgt
• preOrder(ltgt) ltgt
• postOrder(ltgt) ltgt (leere Liste)
• inOrder(ltL,x,Rgt) inOrder(L) ltxgt inOrder(R)
• preOrder(ltL,x,Rgt) ltxgt preOrder(L)
preOrder(R)
• postOrder(ltL,x,Rgt) postOrder(L) postOrder(R)
ltxgt
• Where "" describes list concatenation

7
Example binary tree for the expression
((12/4)2)
• inOrder 12, /, 4, , 2
• preOrder , /, 12, 4, 2
• postOrder 12, 4, /, 2,

8
• Implementation in Java
• public class BinTree
• private Object val
• private BinTree right
• private BinTree left
• // Constructors
• BinTree(Object x)
• val x left right null
• BinTree(Object x)
• val x left right null
• BinTree(Object x, BinTree LTree, BinTree RTree)
• val x left LTree right RTree

9
• // Basic methods (accordirg to signature)
• public boolean isLeaf()
• return ( this.left null this.right
null )
• public Object nodeVal() // according.
"rootEl"
• return this.val
• public void setNodeVal(Object x)
• this.val x
• public BinTree leftTree()
• return this.left
• public BinTree rightTree()
• return this.right
• public static boolean isEmpty(BinTree T)
• return ( T null )

10
• // traversing
• public static LiLiS preOrder(BinTree T)   if (
isEmpty(T) )       return LiLiS.emptyList()
else       return conc3(list1(T.nodeVal()),
preOrder(T.leftTree()),
preOrder(T.rightTree()) )
• Etc.
• // helping (internal) methods for LiLiS-Objekte
• private static LiLiS conc3(LiLiS L1, LiLiS L2,
LiLiS L3)
• return LiLiS.concat(LiLiS.concat(L1,L2),L3)

• private static LiLiS list1(Object el)
• PCell Cel new PCell(el)
• return new LiLiS(Cel)

11
Array-representation
• Beside an implementation with pointers we can
represent a tree directly in an array
• For left-complete binary trees node content in
the following order in the array levels from up
to down an inside each level frpm left to right.
•
• Nodes with index i
• Successor to the right Index 2 i.
• Successor to the left Index 2 i 1.
• parent Index i div 2.

12
Example for array representation
• Note we can also use this for not complete
trees, but in thst case we will have emply places
in the array.

13
Generalization for not binary trees
• A possible definition
• Definition a treeT is a tuple
• T ( x, T1 , ... ,Tk ), where x is a valid
content for the node and and Ti are trees.
• Here k 0 is also valid. The corresponding
trivial tree is composed by only one node.
• (However with this approach belongs a null not
to the set of trees!)

14
Descriptions and Characteristics
• Grade of a node number of children.
• Grade of a tree T
• The maximal number of elements in a tree of
height h and grade d is N(h) (d h1 - 1) / (d
- 1).

15
Implementation
• With an array of pointers to the children (only
when the grade is bounded and the number of
children is not too high).
• Through a pointer to a list of binary trees a
node has a pointer to the leftmost child and to
the sibling next to the right, for example.
• class TreeNode
• private Object val
• private TreeNode leftmostChild
• private TreeNode rightSibling
• ...

16
6.2    Search trees
• Dictionary-Operations
• member
• insert
• delete
• Goal implement these efficiently

17
For comparing
Ordered list (pointer) Non-ordered list (pointer) Ordered list as array
insert O(n) O(1) O(n)
delete O(n) O(n) O(n)
member O(n) O(n) O(log n)
• member in ordered list as array binary search

18
• Definition
• Be T a binary tree.
• N(T) describes the set of nodes of T.
• A mapping N(T) ? D is said to be a node marking
function, where D a range is of complete ordered
values.
• A binary tree T with node marking m is called a
search tree when for each sub-tree
• T ' ( L, x, R ) in T the following holds
• y from L ? m(y) lt m(x)
• y from R ? m(y) gt m(x)
• Note all marks are different (Dictionary!).

19
Implementation
• We derive the BinSearchTree class from the
BinTree class.
• Extension (from the BinTree) marking funktion
• numVal BinSearchTree ? int
• Dictionary-Operations to be implemented
•   ops
• member El x BinSearchTree ? boolean insert
El x BinSearchTree ? BinSearchTree delete El X
BinSearchTree ? BinSearchTree

20
Insert
• Algorithm insert(object x, tree T)
• if T empty then
• makeTree(empty,x,empty return
• if ( m(x) lt m(root of T) ) then
• insert(x, left subtree of T)
• else
• insert(x, left subtree of T)

21
Member
• algorithm member(Object x, treeT)
• if T emply return false
• int k m(x)
• int k m(root of T)
• if k k return true
• if k lt k
• return member(x, left sub-tree of T)
• sonst
• return member(x, right sub-tree ofT)

22
Delete
• delete an element
• (may imply restructuring the tree.)
• 1. Search for the sub-tree T ', whose root has
the element to be deleted.
• 2. If T ' is a leaf, replace T ' in T by null.
• 3. If T ' has only one sub-tree ( T '' ), replace
T ' by T ''.
• 4. Else, delete the smallest element (min) from
the sub-tree at the right of T ' (note min has
at most one sub-tree) and put T '.val min.
(alternatively, delete the node with the biggest
element from the left tree of T ' max and put T '
.val max)

23
Complexity analysis
• Be n the number of nodes in the search tree.
• Costs of a complete traversal O(n)
• Costs for member, insert, delete
• not constant portion search for the right
position in the binary tree across the path
starting from the root
• ? O(height of the tree)

24
• Best case complete tree
• height O(log n).
• Thus the complexity of the operations is only
• O(log n).
• Worst case lineal tree (results when inserting
the elements in order)
• height n-1.
• complexity of the operations
• for each operation O(n),
• for the for the construction of the tree
through inserting the elements O(n²).

25
Complexity analysis (2)
• Average case
• Complexity of the operations are the order of
the average lenght of the path (average of all
paths in all searching trees having n nodes)
• O(log n)
• (siehe Skriptum direkte Abschätzung oder
Berechnung über die Harmonische Reihe)

26
Gesucht Mittlere Astlänge in einem
durchschnittlichen Suchbaum. Dieser sei nur
durch Einfügungen entstanden. Einfügereihenfolge
Alle Permutationen der Menge der Schlüssel a1,
... , an gleichwahrscheinlich. Diese wollen wir
zunächst als sortiert annehmen. A(n) 1
mittlere Astlänge im Baum mit n Schlüsseln A(n)
mittlere Zahl von Knoten auf Pfad in Baum mit
n Schlüsseln. Sei ai1 das erste gewählte
Element. Dann steht dieses Element in der Wurzel.
Im linken Teilbaum finden sich i, im rechten
n-i-1 Elemente. Linker Teilbaum ist zufälliger
Baum mit den Schlüsseln a1 bis ai , rechter mit
den Schlüsseln ai2, ..., an. Die mittlere Zahl
von Knoten auf einem Pfad in diesem Baum ist
daher       i/n ( A(i) 1) (n-i-1)/n
(A(n-i-1) 1) 1(1/n).
27
Dabei sind die Beiträge der Teilbäume um eins
vergrößert (für die Wurzel) und mit den
entsprechenden Gewichten belegt. Der letzte Term
betrifft den Anteil der Wurzel. Schließlich muss
über alle möglichen Wahlen von i mit 0 i lt n
gemittelt werden. So erhalten wir     A(n) n-2
? 0 i lt n i (A(i)1) (n-i-1)(A(n-i-1)1)
1 Aus Symmetriegründen ist der Anteil der
beiden Terme A(i) und A(n-i-1) gleich, die
konstanten Teile summieren sich zu n und
2(n-1)n/2 auf. Somit folgt     A(n) n-2( 2
?0iltn i A(i) (n-1)n n) 1 2n-2?0iltn i
A(i) Wir führen die Abkürzung S(n) ?0iltn i
A(i) ein und erhalten         A(n) 1 2n-2
S(n-1) und         S(n) - S(n-1)  n A(n)   n
2 S(n-1) / n , also die Rekursionsformel
S(n) n S(n-1) (n2) / n.
28
Rekursionsformel S(n) n S(n-1) (n2) /
n Außerdem ist S(0) 0 und S(1) A(1) 1.
Nun wollen wir durch Induktion folgende
Ungleichung beweisen     S(n) n (n1) ln
(n1)  Sicherlich ist dies für  n 0 und n 1
richtig. Einsetzen der Rekursionsformel für n-1
ergibt aber beim Schluss von n-1 nach n
S(n) n S(n-1) (n2) / n
n(n-1) (n2) ln n             n(n1) ln
(n1) (n1)n (ln n - ln (n1)) - 2 ln n n
n(n1) ln (n1) - (n1) n / (n1)
- 2 ln n n             lt n (n1) ln (n1)
Dabei haben wir ln n - ln (n1) -1/(n?),
0lt?lt1, verwendet. Dann folgt aber     A(n) 1
2n-2 S(n-1) 1 2 ln n O( log n)
29
Landis, 1962)
• In normal search trees, the complexity of find,
insert and delete operations in search trees is
in the worst case ?(n).
• Can be better! Idea Balanced trees.
• Definition An AVL-tree is a binary search tree
such that for each sub-tree T ' lt L, x, R gt
h(L) - h(R) ? 1 holds
• (balanced sub-trees is a characteristic of
AVL-trees).
• The balance factor or height is often annotated
at each node h(.)1.

30
Height(I) hight(D) lt 1
This is an AVL tree
31
This is NOT an AVL tree (node does not hold
the required condition)
32
Goals
• 1. How can the AVL-characteristics be kept when
inserting and deleting nodes?
• 2. We will see that for AVL-trees the complexity
of the operations is in the worst case
• O(height of the AVL-tree)
• O(log n)

33
Preservation of the AVL-characteristics
• After inserting and deleting nodes from a tree we
must procure that new tree preserves the
characteristics of an AVL-tree
• Re-balancing.
• How ? simple and double rotations

34
Only 2 cases (an their mirrors)
• Lets analyze the case of insertion
• The new element is inserted at the right (left)
sub-tree of the right (left) child which was
already higher than the left (right) sub-tree by
1
• The new element is inserted at the left (right)
sub-tree of the right (left) child which was
already higher than the left (right) sub-tree by
1

35
Rotation (for the case when the right sub-tree
grows too high after an insertion)
Is transformed into
36
Double rotation (for the case that the right
sub-tree grows too high after an insertion at its
left sub-tree)
Double rotation
Is transformed into
37
b
First rotation
a
c
W
Z
x
y
new
a
Second rotation
b
W
c
x
new
y
Z
38
Re-balancing after insertion
• After an insertion the tree might be still
balanced or
• theorem After an insertion we need only one
rotation of double-rotation at the first node
that got unbalanced in order to re-establish
the balance properties of the AVL tree.
• ( on the way from the inserted node to the
root).
• Because after a rotation or double rotation the
resulting tree will have the original size of the
tree!

39
The same applies for deleting
• Only 2 cases (an their mirrors)
• The element is deleted at the right (left)
sub-tree of which was already smaller than the
left (right) sub-tree by 1
• The new element is inserted at the left (right)
sub-tree of the right (left) child which was
already higher that the left (right) sub-tree by
1

40
The cases
Deleted node
1
1
1
41
Re-balancing after deleting
• After deleting a node the tree might be still
balanced or
• Theorem after deleting we can restore the AVL
balance properties of the sub-tree having as root
the first node that got unbalanced with just
only one simple rotation or a double rotation.
• ( on the way from the deleted note to the
root).
• However the height of the resulting sub-tree
might be shortened by 1, this means more
rotations might be (recursively) necessary at the
parent nodes, which can affect up to the root of
the entire tree.

42
• While searching for unbalanced sub-tree after an
operation it is only necessary to check the
parents sub-tree only when the sons sub-tree
has changed it height.
• In order make the checking for unbalanced
sub-trees more efficient, it is recommended to
example the height of the sub-tree or the
balance factor (height(left sub-tree)
height(right sub-tree)) This information must be
updated after each operation
• It is necessary to have an operation that returns
the parent of a certain node (for example, by
adding a pointer to the parent).

43
Complexity analysis worst case
• Be h the height of the AVL-tree.
• Searching as in the normal binary search tree
O(h).
• Insert the insertion is the same as the binary
search tree (O(h)) but we must add the cost of
one simple or double rotation, which is constant
also O(h).
• delete delete as in the binary search tree(O(h))
but we must add the cost of (possibly) one
rotation at each node on the way from the deleted
node to the root, which is at most the height of
the tree O(h).
• All operations are O(h).

44
Calculating the height of an AVL tree
• Be N(h) the minimal number of nodes
• In an AVL-tree having height h.
• N(0)1, N(1)2,
• N(h) 1 N(h-1) N(h-2) für h ? 2.
• N(3)4, N(4)7
• remember Fibonacci-numbers
• fibo(0)0, fibo(1)1,
• fibo(n) fibo(n-1) fibo(n-2)
• fib(3)1, fib(4)2, fib(5)3, fib(6)5, fib(7)8
• By calculating we can state
• N(h) fibo(h3) - 1

Principle of construction
0
1
2
3
45
• Be n the number of nodes of an AVL-tree of height
h. Then it holds that
• n ? N(h) ,
• By making p (1 sqrt(5))/2 und q (1-
sqrt(5))/2
• we can now write
• n ? fibo(h3)-1
• ( ph3 qh3 ) / sqrt(5) 1
• ? ( p h3/sqrt(5)) 3/2,
• thus
• h3logp(1/sqrt(5)) ? logp(n3/2),
• thus there is a constant c with
• h ? logp(n) c
• logp(2) log2(n) c
• 1.44 log2(n) c O(log n).

46
6.3.1 Heapsort
• Idea two phases
• 1. Construction of the heap
• 2. Output of the heap
• For ordering number in an ascending sequence use
a Heap with reverse order the maximum number
should be at the root (not the minimum).
• Heapsort is an in-situ-Procedure

47
Remembering Heaps change the definition
• Heap with reverse order
• For each node x and each successor y of x the
following holds m(x) ? m(y),
• left-complete, which means the levels are filled
starting from the root and each level from left
to right,
• Implementation in an array, where the nodes are
set in this order (from left to right) .

48
Second Phase

Heap
Heap
Ordered elements
Ordered
elements
• 2. Output of the heap
• take n-times the maximum (in the root,
deletemax)
• and exchange it with the element at the end
of the heap.
• ? Heap is reduced by one element and the
subsequence of ordered elements at the end of the
array grows one element longer.
• cost O(n log n).

49
First Phase
• 1. Construction of the Heap
• simple method n-times insert
• Cost O(n log n).
• making it better consider the array a1 n
as an already left-complete binary tree and
let sink the elements in the following sequence !
• an div 2 a2 a1
• (The elements an an div 2 1 are

HH The
leafs of the heap
50
• Formally heap segment
• an array segment a i..k ( 1 ? i ? k ltn ) is
said to be a heap segment when following holds
•         for all j from i,...,k
• m(a j ) ? m(a 2j )     if 2j ? k and
• m(a j ) ? m(a 2j1)  if 2j1 ? k
• If ai1..n is already a heap segment we can
convert ain into a heap segment by letting
ai sink.

51
Cost calculation
• Be k log n1 - 1. (the height of the
complete portion of the heap)
• cost
• For an element at level j from the root k j.
• alltogether ?j0,,k (k-j)2j
• 2k ?i0,,k i/2i 2 2k O(n).

52
• The new construction strategy is more efficient !
• Usage when only the m biggest elements are
required
• 1. construction in O(n) steps.
• 2. output of the m biggest elements in
• O(mlog n) steps.
• total cost O( n mlog n).

53