递归生成目录树的效率问题 - 专家坐堂 - Foxtable(狐表) - 新一代数据库软件，完美融合Access、Foxpro、Excel、vb.net之优势，人人都能掌握的快速软件开发工具!

以下是引用有点甜在2014/11/27 17:21:00的发言：
'---在各表名下生成目录树节点
e.Sender.StopRedraw
Dim pnd As WinForm.TreeNode = e.Node
If e.Node.level = 1 Then '如果是表名节点
    e.Node.Nodes.Clear '清空原有节点
    Dim tbName As String = e.Node.Name
    Dim dr As DataRow = DataTables("表B").Find("表名 = '" & tbName & "'")
    If dr IsNot Nothing Then
        Dim Cols As String = dr("目录树列")
        Dim Cols_sort As String = Cols.replace("|",",")

        Dim nds As new List(Of WinForm.TreeNode)
        For i As Integer = 0 To Cols.Split("|").Length
            nds.Add(e.Node)
        Next
        If nds.Count > 2 Then
            For Each arys As String() In DataTables(tbName).GetValues(Cols, "", Cols_sort)
                Dim flag As Boolean = False
                For i As Integer = 0 To arys.Length - 1
                    If nds(i+1).Name <> arys(i) OrElse flag Then
                        flag = True
                        nds(i+1) = nds(i).Nodes.Add(arys(i))
                    End If
                Next
            Next
        Else
            For Each arys As String In DataTables(tbName).GetValues(Cols, "", Cols_sort)
                nds(1) = nds(0).Nodes.Add(arys)
            Next
        End If

        e.Node.Expand
    End If
End If
e.Sender.ResumeRedraw

请问这个怎么写到窗口的afterload里实现打开窗口就自动展开全部节点，不是现在需要点击节点才能打开。

主题：递归生成目录树的效率问题