Red black tree sort: Difference between revisions
Red black tree sort in FreeBASIC
(added Phix) |
(Red black tree sort in FreeBASIC) |
||
Line 9:
You may use an implementation at [[Algebraic data types]] as a starting point, if you find that helpful.
=={{header|FreeBASIC}}==
Code originally by AGS.
https://www.freebasic.net/forum/viewtopic.php?t=16113
<lang freebasic>#define NULL Cast(Any Ptr,0)
Enum nodecolor
BLACK = 0
RED = 1
End Enum
Type RBNode
Dim izda As RBNode Ptr
Dim dcha As RBNode Ptr
Dim parent As RBNode Ptr
Dim kolor As nodecolor
Dim key As Integer
Dim value As String
Dim nonzero As Integer
Declare Constructor(Byval key As Integer = 0, value As String = "", Byval clr As nodecolor = RED)
Declare Destructor()
End Type
Constructor RBNode(Byval key As Integer, value As String, Byval clr As nodecolor = RED)
This.key = key
This.value = value
This.izda = NULL
This.dcha = NULL
This.parent = NULL
This.kolor = clr
This.nonzero = 1
End Constructor
Destructor RBNode()
End Destructor
Function integer_compare(Byval key1 As Integer, Byval key2 As Integer) As Integer
If (key1 = key2) Then
Return 0
Elseif (key1 < key2) Then
Return -1
Elseif (key1 > key2) Then
Return 1
End If
End Function
Type RBTree
Dim sentinel As RBNode Ptr
Dim root As RBNode Ptr
Dim count As Integer
Dim Compare As Function(Byval key1 As Integer, Byval key2 As Integer) As Integer
Declare Constructor(Byval cmp As Function(Byval key1 As Integer, Byval key2 As Integer) As Integer)
Declare Sub rotateLeft(Byval x As RBNode Ptr)
Declare Sub rotateRight(Byval x As RBNode Ptr)
Declare Sub insertFixup(Byval x As RBNode Ptr)
Declare Function insertNode(Byval key As Integer, value As String) As RBNode Ptr
Declare Sub deleteFixup(Byval x As RBNode Ptr)
Declare Sub deleteNode(Byval z As RBNode Ptr)
Declare Function findNode(Byval key As Integer) As RBNode Ptr
Declare Destructor()
End Type
Constructor RBTree(Byval cmp As Function(Byval key1 As Integer, Byval key2 As Integer) As Integer)
This.sentinel = New RBNode(0,"",BLACK)
This.sentinel->izda = sentinel
This.sentinel->dcha = sentinel
This.root = This.sentinel
This.count = 0
This.Compare = cmp
End Constructor
Destructor RBTree()
'The tree is transformed into a tree in which
'left children are always leaves. This is done by rotation.
'After rotating any left child is a leaf (not a tree)
'so a izda child can simply be deleted.
'Usually a stack is used to keep track of what nodes have
'been removed. By using rotation there is no need for a stack.
Dim parent As RBNode Ptr
Dim child As RBNode Ptr
If (This.root <> This.sentinel Andalso This.root <> NULL) Then
parent = This.root
While (parent <> This.sentinel)
If (parent->izda = This.sentinel) Then
child = parent->dcha
Delete parent
parent = 0
Else
'rotate
child = parent->izda
parent->izda = child->dcha
child->dcha = parent
End If
parent = child
Wend
Else
If (This.sentinel <> 0) Then
Delete This.sentinel
This.sentinel = 0
End If
End If
End Destructor
Sub RBTree.rotateLeft(Byval x As RBNode Ptr)
'rotate node x to right
Var y = x->dcha
'establish x->dcha link
x->dcha = y->izda
If (y->izda <> This.sentinel) Then
y->izda->parent = x
End If
'establish y->parent link
If (y <> This.sentinel) Then
y->parent = x->parent
End If
If (x->parent) Then
If (x = x->parent->izda) Then
x->parent->izda = y
Else
x->parent->dcha = y
End If
Else
This.root = y
End If
'link x and y
y->izda = x
If (x <> This.sentinel) Then
x->parent = y
End If
End Sub
Sub RBTree.rotateRight(Byval x As RBNode Ptr)
'rotate node x to right
Var y = x->izda
' establish x->left link
x->izda = y->dcha
If (y->dcha <> This.sentinel) Then
y->dcha->parent = x
End If
' establish y->parent link
If (y <> This.sentinel) Then
y->parent = x->parent
End If
If (x->parent) Then
If (x = x->parent->dcha) Then
x->parent->dcha = y
Else
x->parent->izda = y
End If
Else
This.root = y
End If
'link x and y
y->dcha = x
If (x <> This.sentinel) Then
x->parent = y
End If
End Sub
Sub RBTree.insertFixup(Byval x As RBNode Ptr)
'maintain tree balance after inserting node x
'check Red-Black properties
While (x <> This.Root Andalso x->parent->kolor = RED)
'we have a violation
If (x->parent = x->parent->parent->izda) Then
Var y = x->parent->parent->dcha
If (y->kolor = RED) Then
'uncle is RED
x->parent->kolor = BLACK
y->kolor = BLACK
x->parent->parent->kolor = RED
x = x->parent->parent
Else
'uncle is BLACK
If (x = x->parent->dcha) Then
'make x a izda child
x = x->parent
This.rotateLeft(x)
End If
'recolor and rotate
x->parent->kolor = BLACK
x->parent->parent->kolor = RED
This.rotateRight(x->parent->parent)
End If
Else
' mirror image of above code
Var y = x->parent->parent->izda
If (y->kolor = RED) Then
' uncle is RED
x->parent->kolor = BLACK
y->kolor = BLACK
x->parent->parent->kolor = RED
x = x->parent->parent
Else
' uncle is BLACK
If (x = x->parent->izda) Then
x = x->parent
This.rotateRight(x)
End If
x->parent->kolor = BLACK
x->parent->parent->kolor = RED
This.rotateLeft(x->parent->parent)
End If
End If
Wend
This.root->kolor = BLACK
End Sub
Function RBTree.insertNode(Byval key As Integer, value As String) As RBNode Ptr
'Insert a node in the RBTree
'find where node belongs
Dim current As RBNode Ptr = This.root
Dim parent As RBNode Ptr
While (current <> This.sentinel)
Var rc = This.Compare(key, current->key)
If (rc = 0) Then Return current
parent = current
If (rc < 0) Then
current = current->izda
Else
current = current->dcha
End If
Wend
' setup new node
Dim x As RBNode Ptr = New RBNode(key, value)
x->izda = This.sentinel
x->dcha = This.sentinel
x->parent = parent
This.count = This.count + 1
' insert node in tree
If (parent) Then
If (This.Compare(key, parent->key) < 0) Then
parent->izda = x
Else
parent->dcha = x
End If
Else
This.root = x
End If
This.insertFixup(x)
Return x
End Function
Sub RBTree.deleteFixup(Byval x As RBNode Ptr)
'maintain tree balance after deleting node x
Dim w As RBNode Ptr
While (x <> This.root Andalso x->kolor = BLACK)
If (x = x->parent->izda) Then
w = x->parent->dcha
If (w->kolor = RED) Then
w->kolor = BLACK
x->parent->kolor = RED
This.rotateLeft(x->parent)
w = x->parent->dcha
End If
If (w->izda->kolor = BLACK And w->dcha->kolor = BLACK) Then
w->kolor = RED
x = x->parent
Else
If (w->dcha->kolor = BLACK) Then
w->izda->kolor = BLACK
w->kolor = RED
This.rotateRight(w)
w = x->parent->dcha
End If
w->kolor = x->parent->kolor
x->parent->kolor = BLACK
w->dcha->kolor = BLACK
This.rotateLeft(x->parent)
x = This.root
End If
Else
w = x->parent->izda
If (w->kolor = RED) Then
w->kolor = BLACK
x->parent->kolor = RED
This.rotateRight(x->parent)
w = x->parent->izda
End If
If (w->dcha->kolor = BLACK And w->izda->kolor = BLACK) Then
w->kolor = RED
x = x->parent
Else
If (w->izda->kolor = BLACK) Then
w->dcha->kolor = BLACK
w->kolor = RED
This.rotateLeft(w)
w = x->parent->izda
End If
w->kolor = x->parent->kolor
x->parent->kolor = BLACK
w->izda->kolor = BLACK
This.rotateRight(x->parent)
x = This.root
End If
End If
Wend
x->kolor = BLACK
End Sub
Sub RBTree.deleteNode(Byval z As RBNode Ptr)
'delete node z from tree
Dim y As RBNode Ptr
Dim x As RBNode Ptr
If (0 = z Orelse z = This.sentinel) Then Return
If (z->izda = This.sentinel Orelse z->dcha = This.sentinel) Then
'y has a This.sentinel node as a child
y = z
Else
'find tree successor with a This.sentinel node as a child
y = z->dcha
While (y->izda <> This.sentinel)
y = y->izda
Wend
End If
'x is y's only child
If (y->izda <> This.sentinel) Then
x = y->izda
Else
x = y->dcha
End If
'remove y from the parent chain
x->parent = y->parent
If (y->parent) Then
If (y = y->parent->izda) Then
y->parent->izda = x
Else
y->parent->dcha = x
End If
Else
This.root = x
End If
If (y <> z) Then
z->key = y->key
z->value = y->value
End If
If (y->kolor = BLACK) Then
This.deleteFixup(x)
End If
Delete y
This.count = This.count - 1
End Sub
Function RBtree.findNode(Byval key As Integer) As RBNode Ptr
'find node with key equal to key
Var current = This.root
While (current <> This.sentinel)
Var rc = This.Compare(key, current->key)
If (rc = 0) Then
Return current
Else
If (rc < 0) Then
current = current->izda
Else
current = current->dcha
End If
End If
Wend
Return 0
End Function
Type GraphicsNode
Dim node As RBNode Ptr
Dim lvl As Ubyte
Dim nxt As GraphicsNode Ptr
Dim prev As GraphicsNode Ptr
Dim x As Uinteger
Dim y As Uinteger
End Type
Type NodeQueue
Dim startx As Integer
Dim starty As Integer
Dim first As GraphicsNode Ptr
Dim last As GraphicsNode Ptr
Dim levels(2 To 11) As Integer => {100,50,25,12,10,10,10,10,10}
Dim count As Integer
Declare Constructor
Declare Destructor
Declare Function Enqueue(Byref item As GraphicsNode Ptr) As Integer
Declare Function Dequeue(Byref item As GraphicsNode Ptr) As GraphicsNode Ptr
Declare Sub PrintNode(Byval item As GraphicsNode Ptr, Byval x As Integer, Byval y As Integer)
Declare Sub PrintTree(Byval tree As RBTree Ptr)
End Type
Constructor NodeQueue()
''Draw first node in the middle of the screen
'(just below the top of the screen)
This.startx = 350
This.starty = 100
This.first = NULL
This.last = NULL
This.count = 1
'800x600, 32 bits kolor
Screen 19,32
color , Rgb(255,255,155)
Cls
WindowTitle "Red black tree sort"
End Constructor
Destructor NodeQueue()
End Destructor
Function NodeQueue.Enqueue(Byref item As GraphicsNode Ptr) As Integer
'Insertion into an empty que
If (This.first = NULL) Then
This.first = item
This.last = item
This.Count += 1
Return 0
Else
Var tmp = This.last
This.last = item
This.last->prev = tmp
tmp->nxt = This.last
This.last->nxt = NULL
This.Count += 1
Return 0
End If
Return -1
End Function
Function NodeQueue.Dequeue(Byref item As GraphicsNode Ptr) As GraphicsNode Ptr
'Dequeueing from an empty queue or a queue with one node
If (This.last = This.first) Then
'Dequeueing from an empty queue
If (This.last = NULL) Then
This.Count -= 1
Return NULL
Else
'Dequeueing from a queue with one node
item->node = This.First->node
item->x = This.First->x
item->y = This.First->y
item->lvl = This.first->lvl
Delete This.first
This.first = NULL
This.last = NULL
This.Count -= 1
Return item
End If
Else
'Dequeueing from a queue with more than one node
Var tmp = This.Last
item->node = This.Last->node
item->x = This.Last->x
item->y = This.Last->y
item->lvl = This.Last->lvl
This.last = This.last->prev
This.last->nxt = NULL
Delete tmp
Return item
End If
Return NULL
End Function
Sub NodeQueue.PrintNode(Byval item As GraphicsNode Ptr, Byval x As Integer, Byval y As Integer)
'Draw a black line from parent node to child node
Line (x,y)-(item->x,item->y), Rgb(0,0,0)
'Draw node (either red or black)
If (item->node->kolor = RED) Then
Circle (item->x,item->y),5, Rgb(255,0,0),,,,F
Else
Circle (item->x,item->y),5, Rgb(0,0,0),,,,F
End If
Draw String (item->x,item->y - 40), Str(item->node->key), Rgb(0,0,0)
Draw String (item->x-8,item->y - 25),"""" & item->node->value & """", Rgb(0,0,0)
End Sub
Sub NodeQueue.PrintTree(Byval tree As RBTree Ptr)
Dim item As GraphicsNode Ptr
Dim current As GraphicsNode Ptr = New GraphicsNode
Dim tmp As GraphicsNode Ptr
Dim lvl As Integer = 1
Dim x As Integer = This.startx
Dim y As Integer = This.starty
'check for empty tree
If (tree->root = tree->sentinel) Then Return
'Start with printing the root
current->node = tree->root
current->x = x
current->y = y
current->lvl = lvl
This.PrintNode(current,x,y)
Do
'Print izda node (position it at izda side of current node)
If (current->node->izda <> tree->sentinel) Then
item = New GraphicsNode
item->lvl = lvl + 1
If (item->lvl <= 9) Then
item->x = x - This.levels(lvl+1)
Else
item->x = x - 10
End If
item->y = y + 50
item->node = current->node->izda
This.PrintNode(item,x,y)
This.Enqueue(item)
End If
'Print dcha node (position it at dcha side of current node
If (current->node->dcha <> tree->sentinel) Then
item = New GraphicsNode
item->lvl = lvl + 1
If (item->lvl <= 9) Then
item->x = x + This.levels(lvl+1)
Else
item->x = x + 10
End If
item->y = y + 50
item->node = current->node->dcha
This.PrintNode(item,x,y)
This.Enqueue(item)
End If
'Continue drawing from first node in the queue
'Nodes in izda tree will be drawn first as these are put in
'the queue first
Var tmp = This.Dequeue(current)
'If count smaller then entire tree has been drawn
If (This.count < 1) Then Exit Do
x = current->x
y = current->y
lvl = current->lvl
Loop
End Sub
Dim x As Integer Ptr
Dim i As Integer
Var tree = New RBTree(@integer_compare)
Open Cons For Output As #1
For i = 0 To 29
Print #1, "Insert "; i
tree->Insertnode(i,Str(i))
Sleep()
Dim print_tree As NodeQueue Ptr
print_tree = New NodeQueue
print_tree->PrintTree(tree)
Delete print_tree
Next i
Print #1, !"\nStarting Deletion after keypress"
Var print_tree = New NodeQueue
print_tree->PrintTree(tree)
Sleep()
Delete print_tree
randomize timer
For i = 0 To 14
Dim as integer j = int(rnd * 15) + int(rnd * 16)
Print #1, "Delete"; j
Var n = tree->FindNode(j)
If (n) Then tree->Deletenode(n)
Sleep()
Dim print_tree As NodeQueue Ptr
print_tree = New NodeQueue
print_tree->PrintTree(tree)
Delete print_tree
Next i
Bsave "FreeBASIC_Red-black-tree_sort.bmp", 0
Print #1, !"\nEnding program after keypress"
Sleep()
Close #1
Delete tree</lang>
{{out}}
[https://www.dropbox.com/s/hbrtaahsd6jmlyl/FreeBASIC_Red-black-tree_sort.bmp?dl=0 FreeBasic Red black tree sort image]
| |||