问题
I want to make a insert function for a avl tree. However the insert function has to be recursive and must be balanced.
I have a method to pivot the tree left PivoterAGauche and a method to pivot the tree right PivoterADroite.
'Pivot left
Private Function PivoterAGauche(leNoeud As NoeudAVL) As NoeudAVL
If leNoeud Is Nothing Then
Return Nothing
ElseIf leNoeud.FilsDroit Is Nothing AndAlso leNoeud.FilsGauche Is Nothing Then
Return leNoeud
ElseIf leNoeud.FilsDroit Is Nothing Then
Return leNoeud
' ElseIf leNoeud.FilsGauche Is Nothing Then
Else 'Le leNoeud.FilsGauche existe.
Dim pivot As NoeudAVL = leNoeud.FilsGauche
leNoeud.FilsGauche = pivot.FilsDroit
pivot.FilsDroit = leNoeud
leNoeud = pivot
Return leNoeud
End If
End Function
'pivot rigth
Private Function PivoterADroite(leNoeud As NoeudAVL) As NoeudAVL
If leNoeud Is Nothing Then
Return Nothing
ElseIf leNoeud.FilsDroit Is Nothing AndAlso leNoeud.FilsGauche Is Nothing Then
Return leNoeud
ElseIf leNoeud.FilsDroit Is Nothing Then
Return leNoeud
' ElseIf leNoeud.FilsGauche Is Nothing Then
Else 'Le leNoeud.FilsGauche existe.
Dim pivot As NoeudAVL = leNoeud.FilsDroit
leNoeud.FilsDroit = pivot.FilsGauche
pivot.FilsGauche = leNoeud
leNoeud = pivot
Return leNoeud
End If
End Function
Here is the insert method I made. I am usure when to use the pivots left and rigth.
Private Function Inserer(leElement As T, leNoeudCourant As NoeudAVL) As NoeudAVL
Dim intBalance As Integer
'If the node does not existes
If leNoeudCourant Is Nothing Then
m_blnOperationOK = True
Return New NoeudAVL(leElement)
'If the node already existes.
ElseIf leElement.CompareTo(leNoeudCourant.Element) = 0 Then
m_blnOperationOK = False
Return leNoeudCourant
ElseIf leElement.CompareTo(leNoeudCourant.Element) < 0 Then
intBalance = Hauteur(leNoeudCourant.FilsGauche) - Hauteur(leNoeudCourant.FilsDroit)
If (intBalance = 2) Then
leNoeudCourant = PivoterAGauche(leNoeudCourant)
End If
leNoeudCourant.FilsGauche = Inserer(leElement, leNoeudCourant.FilsGauche)
ElseIf leElement.CompareTo(leNoeudCourant.Element) > 0 Then
intBalance = Hauteur(leNoeudCourant.FilsGauche) - Hauteur(leNoeudCourant.FilsDroit)
If (intBalance = 2) Then
leNoeudCourant = PivoterADroite(leNoeudCourant)
End If
leNoeudCourant.FilsDroit = Inserer(leElement, leNoeudCourant.FilsDroit)
End If
'Return current node that will become the root.
Return leNoeudCourant
If you have more questions I will gladly answer your questions thank you for your help.
回答1:
Let's remember something very basic: when using recursion in a task like a binary tree, your workspace is one node. It means that this one node must own everything it needs to be able to do the task you want it to do.
You have to understand that, from it's own point of view, every node is the root of it's own tree, a little bit like if the whole binary tree was built of many many smaller binary trees. O node knows it's children, but not it's ancestors.
To be able to build a balanced binary tree, your nodes must be able to "know" these things:
- Their height (how many nodes are underneath them).
- If they are becoming unbalanced (when, under the currently evaluated node, both children have a height difference higher than 1 level).
The magic here is formidable: once implemented right, the binary tree will always keep it's balance without having to re-evaluate itself entirely. This is big brain time.
First, let's do the height thing:
Every node must have a height value. This value must be stored as a modal variable inside the node class. Make it a public property, because the node's parent will have to be able to access it.
Private _height As Integer = 0
Public ReadOnly Property Height As Integer
Get
Return _height
End Get
End Property
When you insert the first node (the root), it's height is zero. There's nothing underneath. This is true for every new node you create.
When you insert another node, you usually "give" the value to be inserted to the root and let recursion do it's magic. Now there's something else to think about during this operation: update the height, and correct the balance.
Private Function insert(ByVal key As Integer, Optional node As Node = Nothing, ) As Node
'if you are creating the root, node is nothing
If (node Is Nothing) Then
Return New Node(key)
End If
'creating new nodes when needed
If (key < node.key) Then
node.FilGauche = insert(key, node.FilGauche)
ElseIf (key > node.key) Then
node.FilDroit = insert(key, node.FilDroit)
Else
Return node
End If
're-evaluating height (accounting for null pointers) and then balancing the tree
node._height = (1 + max(If(node.FilGauche.Height IsNot Nothing, node.FilGauche.Height, 0), node.FilDroit.Height IsNot Nothing, node.FilDroit.Height))
Dim balance As Integer = If(node.FilGauche.Height IsNot Nothing, node.FilGauche.Height, 0) - If(node.FilDroit.Height IsNot Nothing, node.FilDroit.Height, 0)
' If this node becomes unbalanced, then there
' are 4 cases Left Left Case
If ((balance > 1) AndAlso (key < node.FilGauche.key)) Then
Return rightRotate(node)
End If
' Right Right Case
If ((balance < -1) AndAlso (key > node.FilDroit.key)) Then
Return leftRotate(node)
End If
' Left Right Case
If ((balance > 1) AndAlso (key > node.FilGauche.key)) Then
node.FilGauche = leftRotate(node.FilGauche)
Return rightRotate(node)
End If
' Right Left Case
If ((balance < -1) AndAlso (key < node.FilDroit.key)) Then
node.FilDroit = rightRotate(node.FilDroit)
Return leftRotate(node)
End If
Return node
End Function
Of course, you'll have to adapt these ideas to your exact situation, but that's not the only thing: I'm working from another langugage's references and out of IDE so I have to type directly into my browser, which means that I may have done some mistakes writing this. Let me know if this makes no sense and I'll double-check in better conditions. Good luck!
来源：https://stackoverflow.com/questions/59238949/avl-tree-how-to-balance-tree-on-insert