如何遍历JTree的每一个节点
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参考技术A 下面的例子中将获取到JTree中的每一个节点并按树状结构打印出来：public void testMain(Object[] args)
//Turn off Log Viewer for this example
setOption(IOptionName.BRING_UP_LOGVIEWER, false);
//Start Classics Java Application
startApp("ClassicsJavaA");
// Frame: ClassicsCD
tree2().waitForExistence();
//Display available test data types available from tree
System.out.println ("Available Tree Data Types: " + tree2().getTestDataTypes());
//Declare variables for tree
ITestDataTree cdTree;
ITestDataTreeNodes cdTreeNodes;
ITestDataTreeNode[] cdTreeNode;
//Variables to hold tree data
cdTree = (ITestDataTree)tree2().getTestData("tree");
cdTreeNodes = cdTree.getTreeNodes();
cdTreeNode = cdTreeNodes.getRootNodes();
//Print out total number of nodes
System.out.println ("Tree Total Node Count: " + cdTreeNodes.getNodeCount());
System.out.println ("Tree Root Node Count : " + cdTreeNodes.getRootNodeCount());
//Iterate through tree branches; this is a recursive method.
for (int i = 0;i<cdTreeNode.length;++i)
showTree(cdTreeNode[i], 0);
//Shut down Classics Java Application
classicsJava(ANY,MAY_EXIT).close();
void showTree(ITestDataTreeNode node, int indent)
//Recursive method to print out tree nodes with proper indenting.
//Determine number of tabs to use - to properly indent tree
int tabCount = ( indent < tabs.length() ? indent :
tabs.length() );
//Print out node name + number of children
System.out.println(tabs.substring(0, tabCount) + node.getNode() + " (" + node.getChildCount() + "children)" );
//Determine if node has children; recursively call this same
//method to print out child nodes.
ITestDataTreeNode[] children = node.getChildren();
int childCount = ( children != null ? children.length : 0 );
for ( int i = 0; i < childCount; ++i )
showTree(children[i], indent+1);
//String of tabs used to indent tree view
final String tabs = "/t/t/t/t/t/t/t/t/t/t/t/t/t/t/t/t/t/t/t/t";
输出：
Available Tree Data Types: selected=选中的树层次结构, tree=树层次结构
Tree Total Node Count: 20
Tree Root Node Count : 1
Composers (5children)
Schubert (3children)
String Quartets Nos. 4 & 14 (0children)
Die schone Mullerin, Op. 25 (0children)
Symphonies Nos. 5 & 9 (0children)
Haydn (3children)
Symphonies Nos. 99 & 101 (0children)
Symphonies Nos. 94 & 98 (0children)
Violin Concertos (0children)
Bach (2children)
Brandenburg Concertos Nos. 1 & 3 (0children)
Violin Concertos (0children)
Beethoven (3children)
Symphony No. 7 (0children)
Symphony No. 9 (0children)
Symphony No. 5 (0children)
Mozart (3children)
Symphony No. 34 (0children)
Symphony in C, No. 41: Jupiter (0children)
Concerto in D for Piano (0children)
关于ITestDataTree 、 ITestDataTreeNodes 、ItestDataTreeNode的使用方法可参考RFT的帮助文档：
ITestDataTree
All Superinterfaces:
ITestData
public interface ITestDataTree
extends ITestData
The base-level tree verification-point interface. This interface represents a property set and entry points to the tree nodes.
Since:
RFT1.0
Method Summary
boolean
getOrdered()
Returns the setting of the Ordered property.
ITestDataTreeNodes
getTreeNodes()
Returns the nodes of a tree structure.
void
setOrdered(boolean ordered)
Sets the tree to be compared in either an ordered or unordered fashion.
void
setTreeNodes(ITestDataTreeNodes treeNodes)
Sets the nodes of a tree structure.本回答被提问者和网友采纳
如何在jtree中搜索特定节点并使该节点展开。？
【中文标题】如何在jtree中搜索特定节点并使该节点展开。？【英文标题】：How to search a particular node in jtree and make that node expanded.? 【发布时间】：2012-01-02 21:10:21 【问题描述】：我有一个有 100 个节点的 jtree。现在我想从该树中搜索特定节点并使该节点扩展..？ 我该如何解决这个问题？
【问题讨论】：
【参考方案1】：我假设您的意思是要查找具有特定字符串的节点，对吧？其他答案解释了使用枚举方法执行此操作的方法......（我相信他们都知道，在现实世界中，您还必须满足多个节点具有所寻求的字符串的可能性，等等）
但可能还有其他更性感的方式来做这件事。例如，如果您将所有节点放在某种collection 中（ArrayList 等*），因为它们被插入到树中（并在它们被删除时删除它们，包括显式删除它们的所有后代）。 . 并且如果您还实现了一些事情，如果两个节点从toString 获得相同的结果（或实现了这样做的Comparator），则它们被视为“相等”，那么您可以轻松弹出实际节点（或节点) 在ArrayList 里面匹配，然后去
tree.expandPath( new TreePath( node_found.getPath())
树的一个要点是它实际上是节点的路径（有时称为“面包屑”），它是任何给定节点的真正“身份”。就显示的String 值而言，这意味着您可能在同一棵树中拥有：
路径：“彼得”-“派珀”-“腌制”-“胡椒” 路径：“烹饪特色”-“香料”-“胡椒” 路径：“我的最爱”-“食物”-“调味品”-“胡椒”
假设您要搜索，然后选择或突出显示这些“胡椒”节点之一……为每个元素采用“蛮力”枚举方法并不是很有效路径（树越大，问题当然越严重）。
使用我的建议，它变得非常简单：只需从根或任何位置开始划分您的“面包屑”路径，然后，当您深入到树中时，在“更高”节点上使用 node.isNodeDescendant() )（即那些离根更远的）你已经找到了（这里是 3 个“胡椒”节点）：如果你想要上面的第一个路径，你会首先找到节点“彼得”，然后紧接着唯一的“胡椒”可以满足isNodeDescendant 测试的节点将产生您正在寻找的整个路径。
* 当然，某种形式的散列集合会更有效。但只有在树中有数千个或更多节点时才需要考虑这一点。
【讨论】：
【参考方案2】：这是一个如何在搜索中循环遍历树的示例：
import java.awt.BorderLayout;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.ArrayList;
import java.util.Enumeration;
import java.util.List;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JScrollPane;
import javax.swing.JTextField;
import javax.swing.JTree;
import javax.swing.tree.DefaultMutableTreeNode;
import javax.swing.tree.DefaultTreeModel;
import javax.swing.tree.TreePath;
public class TreeDemo extends JFrame implements ActionListener
private static final long serialVersionUID = 1L;
public JTree tree;
public JButton button;
public JTextField text;
public TreeDemo()
button = new JButton("Enter search text below and click");
text = new JTextField();
button.addActionListener(this);
tree = new JTree();
DefaultMutableTreeNode root = new DefaultMutableTreeNode( "Deck" );
DefaultMutableTreeNode itemClubs= new DefaultMutableTreeNode( "Clubs" );
addAllCard( itemClubs );
root.add( itemClubs );
DefaultMutableTreeNode itemDiamonds = new DefaultMutableTreeNode( "Diamonds" );
addAllCard( itemDiamonds );
root.add( itemDiamonds );
DefaultMutableTreeNode itemSpades = new DefaultMutableTreeNode( "Spades" );
addAllCard( itemSpades );
root.add( itemSpades );
DefaultMutableTreeNode itemHearts = new DefaultMutableTreeNode( "Hearts" );
addAllCard( itemHearts );
root.add( itemHearts );
DefaultTreeModel treeModel = new DefaultTreeModel( root );
tree = new JTree( treeModel );
JScrollPane scrollPane = new JScrollPane(tree);
getContentPane().add(scrollPane, BorderLayout.CENTER);
getContentPane().add(button, BorderLayout.NORTH);
getContentPane().add(text, BorderLayout.SOUTH);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
setSize(375, 400);
public void addAllCard( DefaultMutableTreeNode suit )
suit.add( new DefaultMutableTreeNode( "Ace" ) );
suit.add( new DefaultMutableTreeNode( "Two" ) );
suit.add( new DefaultMutableTreeNode( "Three" ) );
suit.add( new DefaultMutableTreeNode( "Four" ) );
suit.add( new DefaultMutableTreeNode( "Five" ) );
suit.add( new DefaultMutableTreeNode( "Six" ) );
suit.add( new DefaultMutableTreeNode( "Seven" ) );
suit.add( new DefaultMutableTreeNode( "Eight" ) );
suit.add( new DefaultMutableTreeNode( "Nine" ) );
suit.add( new DefaultMutableTreeNode( "Ten" ) );
suit.add( new DefaultMutableTreeNode( "Jack" ) );
suit.add( new DefaultMutableTreeNode( "Queen" ) );
suit.add( new DefaultMutableTreeNode( "King" ) );
public final DefaultMutableTreeNode findNode(String searchString)
List<DefaultMutableTreeNode> searchNodes = getSearchNodes((DefaultMutableTreeNode)tree.getModel().getRoot());
DefaultMutableTreeNode currentNode = (DefaultMutableTreeNode)tree.getLastSelectedPathComponent();
DefaultMutableTreeNode foundNode = null;
int bookmark = -1;
if( currentNode != null )
for(int index = 0; index < searchNodes.size(); index++)
if( searchNodes.get(index) == currentNode )
bookmark = index;
break;
for(int index = bookmark + 1; index < searchNodes.size(); index++)
if(searchNodes.get(index).toString().toLowerCase().contains(searchString.toLowerCase()))
foundNode = searchNodes.get(index);
break;
if( foundNode == null )
for(int index = 0; index <= bookmark; index++)
if(searchNodes.get(index).toString().toLowerCase().contains(searchString.toLowerCase()))
foundNode = searchNodes.get(index);
break;
return foundNode;
private final List<DefaultMutableTreeNode> getSearchNodes(DefaultMutableTreeNode root)
List<DefaultMutableTreeNode> searchNodes = new ArrayList<DefaultMutableTreeNode>();
Enumeration<?> e = root.preorderEnumeration();
while(e.hasMoreElements())
searchNodes.add((DefaultMutableTreeNode)e.nextElement());
return searchNodes;
public static void main(String[] args)
TreeDemo app = new TreeDemo();
app.setVisible(true);
public void actionPerformed(ActionEvent e)
String search = text.getText();
if(search.trim().length() > 0 )
DefaultMutableTreeNode node = findNode(search);
if( node != null )
TreePath path = new TreePath(node.getPath());
tree.setSelectionPath(path);
tree.scrollPathToVisible(path);
【讨论】：
好例子；也可以考虑text.addActionListener(this)；另见Initial Threads。【参考方案3】：
扩展@mKorbel 的答案并如How to Use Trees 中所述，您可以递归搜索TreeModel 并获得TreePath 到结果节点。一旦你有了想要的path，就很容易在树中显示它。
tree.setSelectionPath(path);
tree.scrollPathToVisible(path);
附录：这是“获取TreePath”的一种方法。
private TreePath find(DefaultMutableTreeNode root, String s)
@SuppressWarnings("unchecked")
Enumeration<DefaultMutableTreeNode> e = root.depthFirstEnumeration();
while (e.hasMoreElements())
DefaultMutableTreeNode node = e.nextElement();
if (node.toString().equalsIgnoreCase(s))
return new TreePath(node.getPath());
return null;
【讨论】：
我的（狂野的 :-) 猜测是“获得 TreePath”是问题的核心——不是火箭科学，但在教程中没有这样做的例子（或者我是盲人, 非零概率以上是关于如何遍历JTree的每一个节点的主要内容，如果未能解决你的问题，请参考以下文章