如果我有一个 TreeView (myTreeview),我如何获得所有作为父节点的节点的列表?即有孩子的节点
TK
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2 回答
5
myTreeview.Nodes 将为您提供 MS 定义的根节点列表,这基本上意味着树根分支上的节点。
此代码将构建一个包含子节点的根节点列表:
IList<TreeNode> nodesWithChildren = new List<TreeNode>();
foreach( TreeNode node in myTreeview.Nodes )
if( node.Nodes.Count > 0 ) nodesWithChildren.Add( node );
更新:如果您希望 TreeView 中的所有节点都有一个子节点,而不管树有多深,那么请使用一些递归,例如
private static IList<TreeNode> BuildParentNodeList(TreeView treeView)
{
IList<TreeNode> nodesWithChildren = new List<TreeNode>();
foreach( TreeNode node in treeView.Nodes )
AddParentNodes(nodesWithChildren, node);
return nodesWithChildren;
}
private static void AddParentNodes(IList<TreeNode> nodesWithChildren, TreeNode parentNode)
{
if (parentNode.Nodes.Count > 0)
{
nodesWithChildren.Add( parentNode );
foreach( TreeNode node in parentNode.Nodes )
AddParentNodes( nodesWithChildren, node );
}
}
更新 2:只有 1 个 foreach 循环的递归方法:
private static IList<TreeNode> BuildParentNodeList(TreeView treeView)
{
IList<TreeNode> nodesWithChildren = new List<TreeNode>();
AddParentNodes( nodesWithChildren, treeView.Nodes );
return nodesWithChildren;
}
private static void AddParentNodes(IList<TreeNode> nodesWithChildren, TreeNodeCollection parentNodes )
{
foreach (TreeNode node in parentNodes)
{
if (node.Nodes.Count > 0)
{
nodesWithChildren.Add( node );
AddParentNodes(nodesWithChildren, node.Nodes);
}
}
}
于 2009-02-18T11:25:03.497 回答
1
private void AddNonLeafNodes(List<TreeNode> nonLeafNodes, TreeNodeCollection nodes)
{
foreach( TreeNode node in nodes )
{
if( node.Nodes.Count > 0 )
{
nonLeafNodes.Add(node);
AddNonLeafNodes(nonLeafNodes,node.Nodes);
}
}
}
List<TreeNode> nonLeafNodes = new List<TreeNode>();
AddNonLeafNodes(nonLeafNodes,treeView1.Nodes);
于 2009-02-18T11:44:56.990 回答