我想创建一个函数如下:
假设我有一个WinForms树视图，看起来如下:
Parent1
Child1
Sub-Child1
DeepestNode1
DeepestNode2
DeepestNode3
Sub-Child2
DeepestNode4
DeepestNode5
DeepestNode6
Child2
Sub-Child3
Sub-Child4
Sub-Child5
Sub-Child6
Child3
(no children)
我想创建一个函数如下:
Function GetDeepestChildren(MyNode as Treenode) as List(Of Treenode)
其中，如果结果看起来像:
GetDeepestChildren(Parent1) = {DeepestNode1, DeepestNode2, DeepestNode3, DeepestNode4, DeepestNode5, DeepestNode6}
GetDeepestChildren(Sub-Child1) = {DeepestNode1, DeepestNode2, DeepestNode3}
GetDeepestChildren(Child2) = {Sub-Child3, Sub-Child4, Sub-Child5, Sub-Child6}
GetDeepestChildren(Child3) = Empty list
…换句话说，始终从给定的节点进入最深层并返回子节点——即使它们在不同的父节点之间被分割(如Parent1中的情况)。
我已经创建了一个函数，它将告诉我一个节点更深了多少层，它看起来像:
Public Function GetDeepestChildNodeLevel(ByVal ParentNode As TreeNode) As Integer
Dim subLevel = ParentNode.Nodes.Cast(Of TreeNode).Select(Function(subNode) GetDeepestChildNodeLevel(subNode))
Return If(subLevel.Count = 0, 0, subLevel.Max() + 1)
End Function
所以我知道从什么级别得到孩子，我要找的是一个函数，可以做到这一点-沿着行的东西:
Function GetDeepestChildren(MyNode as Treenode) as List(Of Treenode)
Return All child nodes where level = GetDeepestChildNodeLevel(MyNode)
End function
我希望这是有意义的-谢谢!
在c#中，您可以使用yield return或递归lambda来完成。下面是第二种方法的例子:
Func<TreeNode,IEnumerable<TreeNode>> getChildren = null;
getChildren = n => {
if (n.Nodes.Count != 0) {
var list = new List<TreeNode>(n.Nodes.Where(c => c.Nodes.Count == 0));
foreach (var c in n.Nodes) {
// Note the recursive call below:
list.AddRange(getChildren(c));
}
return list;
} else {
return new TreeNode[0];
}
};
var res = getChildren(myTree);
这是一个使用XML的版本——翻译应该很容易。我用的是我推荐的linkPad，你可以直接在linkPad
中运行它WalkDeep(tree,getDeep(tree)) returns:
<DeepestNode1 />
<DeepestNode2 />
<DeepestNode3 />
<DeepestNode4 />
<DeepestNode5 />
<DeepestNode6 />
function getDeep( e as XElement) as integer
if (e.HasElements)
return 1 + e.Elements().Select(Function(c) getDeep(c)).Max()
else
return 1
end if
end function
function WalkDeep(root as XElement,find as integer,optional mylevel as integer = 1) as IEnumerable(of XElement)
Dim result As New List(Of XElement)
if find = mylevel
result.Add(root)
else
if root.HasElements
for each c as XElement in root.Elements()
for each r as XElement in WalkDeep(c,find,mylevel+1)
result.Add(r)
next
next
end if
end if
return result
end function
Sub Main
dim tree as XElement = <Parent1>
<Child1>
<Sub-Child1>
<DeepestNode1/>
<DeepestNode2/>
<DeepestNode3/>
</Sub-Child1>
<Sub-Child2>
<DeepestNode4/>
<DeepestNode5/>
<DeepestNode6/>
</Sub-Child2>
</Child1>
<Child2>
<Sub-Child3/>
<Sub-Child4/>
<Sub-Child5/>
<Sub-Child6/>
</Child2>
<Child3 />
</Parent1>
WalkDeep(tree,getDeep(tree)).Select(function(x) x.Name.LocalName).Dump()
End Sub
int getDeep(XElement e)
{
if (e.HasElements)
return 1 + e.Elements().Select(c => getDeep(c)).Max();
else
return 1;
}
IEnumerable<XElement> WalkDeep(XElement root,int find, int mylevel=1)
{
if (find == mylevel) yield return root;
if (root.HasElements)
{
foreach(XElement c in root.Elements())
{
foreach(XElement r in WalkDeep(c,find,mylevel+1))
yield return r;
}
}
yield break;
}
void Main()
{
XElement tree = XElement.Parse (@"
<Parent1>
<Child1>
<Sub-Child1>
<DeepestNode1/>
<DeepestNode2/>
<DeepestNode3/>
</Sub-Child1>
<Sub-Child2>
<DeepestNode4/>
<DeepestNode5/>
<DeepestNode6/>
</Sub-Child2>
</Child1>
<Child2>
<Sub-Child3/>
<Sub-Child4/>
<Sub-Child5/>
<Sub-Child6/>
</Child2>
<Child3 />
</Parent1>
");
WalkDeep(tree,getDeep(tree)).Dump();
}
这是一个VB程序。Net重新制作我创建的@dasblinkenlight的解决方案-它工作得很好，我只是把它放在这里以防将来有人需要VB的解决方案。
Public Function GetDeepestChildNodes(ByVal ParentNode As TreeNode) As List(Of TreeNode)
Dim RetVal As New List(Of TreeNode)
If ParentNode.Nodes.Count > 0 Then
RetVal = (From nd As TreeNode In ParentNode.Nodes
Where nd.Nodes.Count = 0
Select nd).ToList
For Each nd In ParentNode.Nodes
RetVal.AddRange(GetDeepestChildNodes(nd))
Next
End If
Return RetVal
End Function
尝试递归函数了吗?在VB中不确定。Net但是c#看起来像
public List<TreeNode> GetDeepestChildren(Treenode MyNode)
{
if (MyNode.Children.Count > 0)
GetDeepestChildren(Treenode MyNode);
else
return MyNode.Children;
}
这还没有被编译或测试，但想法是存在的，应该返回任何给定节点的最深处的子节点。
我不熟悉TreeView控件，但是TreeNode的Level属性对你有什么好处吗?
http://msdn.microsoft.com/en-us/library/system.windows.forms.treenode.level.aspx如果你知道最深的层次，你可以这样做:
c#private List<TreeNode> GetDeepestChildren(int level)
{
return (from p in treeView1.Nodes.Cast<TreeNode>() where p.Level == level select p).ToList();
}
VB
Private Function GetDeepestChildren(level As Integer) As List(Of TreeNode)
Return (From p In treeView1.Nodes.Cast(Of TreeNode)() Where p.Level = levelp).ToList()
End Function
这是只是对@dasblinkenlight的回答的轻微修改，所以不要给这个投票!
只是个人风格的问题，但我更喜欢这样的递归调用:
IEnumerable<TreeNode> WalkNodes(TreeNode root)
{
yield return root;
var children = root.Nodes.Cast<TreeNode>();
foreach (var child in children)
{
foreach(var subChild in WalkNodes(child))
{
yield return subChild;
}
}
}
通过
调用foreach (var node in treeView.Nodes.Cast<TreeNode>())
{
var walkedFrom = WalkNodes(node);
foreach (var subNode in walkedFrom)
{
Console.WriteLine(subNode.Text);
}
}
它不是linq，因为我认为在这种情况下，它不应该由linq完成抱歉，如果它不是你问的，而是这个工作。它不是完全证明的，但至少你不会得到stackoverflow如果你得到一个疯狂的树
Private Sub Form1_Load(sender As System.Object, e As System.EventArgs) Handles MyBase.Load
Dim test1 = GetDeepestChildren(TreeView1.Nodes(0))
Dim test2 = GetDeepestChildren(TreeView1.Nodes(0).Nodes(0).Nodes(0))
Dim test3 = GetDeepestChildren(TreeView1.Nodes(0).Nodes(1))
Dim test4 = GetDeepestChildren(TreeView1.Nodes(0).Nodes(2))
End Sub
Private Function GetDeepestChildren(ByVal node As TreeNode) As List(Of TreeNode)
Dim deepestList As New List(Of TreeNode)
If node.Nodes.Count = 0 Then
Return deepestList
End If
Dim nodes As New Stack(Of TreeNode)
For Each n As TreeNode In node.Nodes
nodes.Push(n)
Next
Dim deepest As Integer = 0
Do Until nodes.Count = 0
node = nodes.Pop
If node.Nodes.Count = 0 Then
If deepest < node.Level Then
deepest = node.Level
deepestList.Clear()
deepestList.Add(node)
ElseIf deepest = node.Level Then
deepestList.Add(node)
End If
Else
For Each n As TreeNode In node.Nodes
nodes.Push(n)
Next
End If
Loop
Return deepestList
End Function