C# TreeView使用,读取XML。

Question

 XML示例：
。
   <info>
   <memu>1</memu>
   <datainfo>demo1</datainfo>
   </info>
   <info>
   <memu>2</memu>
   <datainfo>demo2</datainfo>
   </info>
   <info>
   <memu>2</memu>
   <datainfo>demo3</datainfo>
   </info>
。。
需求描述：
  1.查出节点为memu的值设为父节点，datainfo的值设置为子节点。
  2.有重复的memu的值就并和到一个父节点下。  生成如下所示树状列：
   1
     demo1
   2 
     demo2
     demo3问题：我已用递归取到每个info内的节点集，但不知如何判断单个节点是否为memu,用SelectSingleNode报错。

Answer 1

 private void XmlOperation_Load(object sender, EventArgs e)
        {
            path = AppDomain.CurrentDomain.BaseDirectory + @"NameList.xml";
            xml.Load(path);//加载xml文件
            bindTvXml();
        }        /// <summary>
        /// 绑定TreeView
        /// </summary>
        private void bindTvXml()
        {
            for (int i = 0; i < xml.DocumentElement.ChildNodes.Count; i++)
            {
                XmlNode Xnode = xml.DocumentElement.ChildNodes[i];
                TreeNode node = new TreeNode();
                node.Text = Xnode.Attributes["name"].Value;
                node.Tag = Xnode;
                bindChildNode(node, Xnode);//绑定子节点
                TvXml.Nodes.Add(node);
                TvXml.HideSelection = false;
            }
        }
        
        /// <summary>
        /// 递归绑定子节点
        /// </summary>
        /// <param name="node"></param>
        /// <param name="xml"></param>
        private void bindChildNode(TreeNode node, XmlNode xml)
        {
            for (int i = 0; i < xml.ChildNodes.Count; i++)
            {
                TreeNode Childnode = new TreeNode();
                XmlNode ChildXml = xml.ChildNodes[i];
                Childnode.Text = ChildXml.Value;
                Childnode.Name = "1";
                Childnode.Tag = xml.ChildNodes[i];
                if (ChildXml.HasChildNodes)
                {
                    if (ChildXml.ChildNodes[0].NodeType == XmlNodeType.Text)
                        Childnode.Text = ChildXml.ChildNodes[0].InnerText;
                    else
                        bindChildNode(Childnode, ChildXml);
                }
                node.Nodes.Add(Childnode);
            }
           
        }

Answer 2

  protected void Page_Load(object sender, EventArgs e)
        {
            string xml = @"<info>
                              <memu>1</memu>
                              <datainfo>demo1</datainfo>
                              </info>
                              <info>
                              <memu>2</memu>
                              <datainfo>demo2</datainfo>
                              </info>
                              <info>
                              <memu>2</memu>
                              <datainfo>demo3</datainfo>
                              </info>";
            XElement x = XElement.Load(Server.MapPath("~/Web.config"));
            foreach (XElement r in x.Elements())
            {
                TreeNode node = new TreeNode();
                node.Text = r.Name.ToString();
                BindChild(node,r.Elements());
                TreeView1.Nodes.Add(node);
            };
        }
        public void BindChild(TreeNode node,IEnumerable<XElement> xs) 
        {
            foreach (XElement r in xs)
            {
                TreeNode ChildNode = new TreeNode();
                ChildNode.Text = r.Name.ToString();
                node.ChildNodes.Add(ChildNode);
                BindChild(ChildNode, r.Elements());
            };
        }LINQ绑TREEVIEW