起始现在好多方法都是先生成XML，然后在生成TreeView。其实，本身创建一个动态的TreeView并不是件很复杂的事。

以下就具体说明，如何产生一个动态的树：

首先，为这个树先要获取数据，如下：

private DataTable dtTree = null;

private DataView dv = null;

private void CreateTable()

{

dtTree = new DataTable( "TreeDBInfo" );

dtTree.Columns.Add( new DataColumn( "TypeID", typeof(int) ) );

dtTree.Columns.Add( new DataColumn( "TypeName", typeof(string) ) );

dtTree.Columns.Add( new DataColumn( "TypeParentID", typeof(int) ) );

dtTree.Rows.Add( new object[]{1, "Parent1", 0} );//'0' is that the type has no parent

dtTree.Rows.Add( new object[]{2, "Parent2", 0} );

dtTree.Rows.Add( new object[]{3, "Parent3", 0} );

dtTree.Rows.Add( new object[]{ 4, "Child1", 1} );

dtTree.Rows.Add( new object[]{ 5, "Child2", 1} );

dtTree.Rows.Add( new object[]{ 6, "Child3", 2} );

dtTree.Rows.Add( new object[]{ 7, "GrandChild1", 4} );

dtTree.Rows.Add( new object[]{ 8, "GrandChild2", 4} );

dtTree.Rows.Add( new object[]{ 9, "GrandChild3", 5} );

}

这是一个简单的属性结构，如果要从数据库读出的话，只要为dtTable指定为已经在DataSet填充好的Table即可，这里就不细说了。

接下来，就是生成树的函数，如下：

private void CreateTree()

{

dv = dtTree.DefaultView;

dv.Sort = "TypeParentID ASC";

DataRowView[] arrDRV = dv.FindRows( 0 );//Get root data info

if( arrDRV.Length == 0 ) return;

TreeNode tnNew = null;

foreach( DataRowView drv in arrDRV )

{

tnNew = trvDBBinding.Nodes.Add( drv.Row["TypeName"].ToString() );

tnNew.Tag = drv.Row["TypeID"].ToString();//Save "TypeID" in node's tag

CreateTreeNode( ref tnNew );

}

}

private void CreateTreeNode( ref TreeNode tnParent )

{

DataRowView[] arrDRV = dv.FindRows( tnParent.Tag );//Get children data info

if( arrDRV.Length == 0 ) return;

TreeNode tnNew = null;

foreach( DataRowView drv in arrDRV )

{

tnNew = tnParent.Nodes.Add( drv.Row["TypeName"].ToString() );

tnNew.Tag = drv.Row["TypeID"].ToString();//Save "TypeID" in node's tag

CreateTreeNode( ref tnNew );

}

}

简单调用如下：

CreateTable();

CreateTree();