Check for foldable binary tree in vb.net

Vb program for Check for foldable binary tree. Here problem description and other solutions.

' Include namespace system
Imports System 
'  Vb.net program for
'  Check if binary tree is foldable binary tree
'  Using recursion

'  Binary Tree Node
Public Class TreeNode
    Public  data As Integer
    Public  left As TreeNode
    Public  right As TreeNode
    Public Sub New(ByVal data As Integer)
        '  Set node value
        Me.data = data
        Me.left = Nothing
        Me.right = Nothing
    End Sub
End Class

public Class BinaryTree
    Public  root As TreeNode
    Public Sub New()
        '  Set initial tree root
        Me.root = Nothing
    End Sub
    Public Function  isFoldable(ByVal leftRoot As TreeNode, 
                                ByVal rightRoot As TreeNode) As Boolean
        if (leftRoot  Is  Nothing AndAlso 
            rightRoot  Is  Nothing) Then
            '  When both node empty
            Return  True
        ElseIf (leftRoot IsNot Nothing AndAlso 
                rightRoot IsNot Nothing AndAlso 
                Me.isFoldable(leftRoot.left, rightRoot.right) AndAlso 
        		Me.isFoldable(leftRoot.right, rightRoot.left)) Then
            '  When
            '  Valid the properties of foldable tree
            Return  True
        End If
        '  Fail case
        Return  False
    End Function
    '  Display tree element inorder form
    Public Sub inorder(ByVal node As TreeNode)
        if (node IsNot Nothing) Then
            '  Visit to left subtree
            Me.inorder(node.left)
            '  Print node value
            Console.Write("  " + node.data.ToString())
            '  Visit to right subtree
            Me.inorder(node.right)
        End If
    End Sub
    Public Shared Sub Main(ByVal args As String())
        '  Create new tree
        Dim tree As BinaryTree = New BinaryTree()
        '   Binary Tree
        '  -----------------------
        '       5
        '     /   \
        '    7     4
        '   /       \
        '  1         2
        '   \       /
        '    2     5
        '  Binary tree nodes
        tree.root = New TreeNode(5)
        tree.root.left = New TreeNode(7)
        tree.root.right = New TreeNode(4)
        tree.root.left.left = New TreeNode(1)
        tree.root.right.right = New TreeNode(2)
        tree.root.right.right.left = New TreeNode(5)
        tree.root.left.left.right = New TreeNode(2)
        '  Display Tree elements
        tree.inorder(tree.root)
        Dim result As Boolean = tree.isFoldable(tree.root.left, 
                                                tree.root.right)
        if (result = True) Then
            Console.WriteLine( vbLf &" Foldable Tree")
        Else
            Console.WriteLine( vbLf &" Not Foldable Tree")
        End IF
        '         5
        '       /   \
        '      7     4
        '     /       \
        '    1         2
        '     \       /
        '      2     5
        '       \
        '        3  
        '  Case 2
        tree.root.left.left.right.right = New TreeNode(3)
        tree.inorder(tree.root)
        result = tree.isFoldable(tree.root.left, tree.root.right)
        if (result = True) Then
            Console.WriteLine( vbLf &" Foldable Tree")
        Else
            Console.WriteLine( vbLf &" Not Foldable Tree")
        End IF
    End Sub
End Class

Output

  1  2  7  5  4  5  2
 Foldable Tree
  1  2  3  7  5  4  5  2
 Not Foldable Tree