我写了一个处理通用树的小库。它比秋千轻多了。我也有一个专门的项目。

首先应该定义什么是树(对于域)，最好先定义接口。并不是所有的树结构都是可修改的，能够添加和删除节点应该是一个可选的功能，所以我们为此做了一个额外的接口。

没有必要创建保存值的节点对象，事实上，我认为这是大多数树实现中的主要设计缺陷和开销。如果查看Swing, TreeModel没有节点类(只有DefaultTreeModel使用TreeNode)，因为实际上并不需要它们。

public interface Tree <N extends Serializable> extends Serializable {
    List<N> getRoots ();
    N getParent (N node);
    List<N> getChildren (N node);
}

可变树结构(允许添加和删除节点):

public interface MutableTree <N extends Serializable> extends Tree<N> {
    boolean add (N parent, N node);
    boolean remove (N node, boolean cascade);
}

有了这些接口，使用树的代码就不必太关心树是如何实现的。这允许您使用通用实现和专用实现，在专用实现中，通过将函数委托给另一个API来实现树。

例如:文件树结构

public class FileTree implements Tree<File> {

    @Override
    public List<File> getRoots() {
        return Arrays.stream(File.listRoots()).collect(Collectors.toList());
    }

    @Override
    public File getParent(File node) {
        return node.getParentFile();
    }

    @Override
    public List<File> getChildren(File node) {
        if (node.isDirectory()) {
            File[] children = node.listFiles();
            if (children != null) {
                return Arrays.stream(children).collect(Collectors.toList());
            }
        }
        return Collections.emptyList();
    }
}

示例:通用树结构(基于父/子关系):

public class MappedTreeStructure<N extends Serializable> implements MutableTree<N> {

    public static void main(String[] args) {

        MutableTree<String> tree = new MappedTreeStructure<>();
        tree.add("A", "B");
        tree.add("A", "C");
        tree.add("C", "D");
        tree.add("E", "A");
        System.out.println(tree);
    }

    private final Map<N, N> nodeParent = new HashMap<>();
    private final LinkedHashSet<N> nodeList = new LinkedHashSet<>();

    private void checkNotNull(N node, String parameterName) {
        if (node == null)
            throw new IllegalArgumentException(parameterName + " must not be null");
    }

    @Override
    public boolean add(N parent, N node) {
        checkNotNull(parent, "parent");
        checkNotNull(node, "node");

        // check for cycles
        N current = parent;
        do {
            if (node.equals(current)) {
                throw new IllegalArgumentException(" node must not be the same or an ancestor of the parent");
            }
        } while ((current = getParent(current)) != null);

        boolean added = nodeList.add(node);
        nodeList.add(parent);
        nodeParent.put(node, parent);
        return added;
    }

    @Override
    public boolean remove(N node, boolean cascade) {
        checkNotNull(node, "node");

        if (!nodeList.contains(node)) {
            return false;
        }
        if (cascade) {
            for (N child : getChildren(node)) {
                remove(child, true);
            }
        } else {
            for (N child : getChildren(node)) {
                nodeParent.remove(child);
            }
        }
        nodeList.remove(node);
        return true;
    }

    @Override
    public List<N> getRoots() {
        return getChildren(null);
    }

    @Override
    public N getParent(N node) {
        checkNotNull(node, "node");
        return nodeParent.get(node);
    }

    @Override
    public List<N> getChildren(N node) {
        List<N> children = new LinkedList<>();
        for (N n : nodeList) {
            N parent = nodeParent.get(n);
            if (node == null && parent == null) {
                children.add(n);
            } else if (node != null && parent != null && parent.equals(node)) {
                children.add(n);
            }
        }
        return children;
    }

    @Override
    public String toString() {
        StringBuilder builder = new StringBuilder();
        dumpNodeStructure(builder, null, "- ");
        return builder.toString();
    }

    private void dumpNodeStructure(StringBuilder builder, N node, String prefix) {
        if (node != null) {
            builder.append(prefix);
            builder.append(node.toString());
            builder.append('\n');
            prefix = "  " + prefix;
        }
        for (N child : getChildren(node)) {
            dumpNodeStructure(builder, child, prefix);
        }
    }
}

与Gareth的答案相同，请检查DefaultMutableTreeNode。它不是一般的，但在其他方面似乎符合要求。即使它在javax中。swing包，它不依赖于任何AWT或swing类。事实上，源代码实际上有注释// ISSUE:这个类不依赖于AWT中的任何东西——移到java.util?

我写了一个基于“HashMap”的小“TreeMap”类，它支持添加路径:

import java.util.HashMap;
import java.util.LinkedList;

public class TreeMap<T> extends LinkedHashMap<T, TreeMap<T>> {

    public void put(T[] path) {
        LinkedList<T> list = new LinkedList<>();
        for (T key : path) {
            list.add(key);
        }
        return put(list);
    }

    public void put(LinkedList<T> path) {
        if (path.isEmpty()) {
            return;
        }
        T key = path.removeFirst();
        TreeMap<T> val = get(key);
        if (val == null) {
            val = new TreeMap<>();
            put(key, val);
        }
        val.put(path);
    }

}

它可以用来存储一个“T”(泛型)类型的树，但(目前)不支持在节点中存储额外的数据。如果你有一个这样的文件:

root, child 1
root, child 1, child 1a
root, child 1, child 1b
root, child 2
root, child 3, child 3a

然后你可以通过执行:

TreeMap<String> root = new TreeMap<>();
Scanner scanner = new Scanner(new File("input.txt"));
while (scanner.hasNextLine()) {
  root.put(scanner.nextLine().split(", "));
}

你会得到一棵漂亮的树。它应该很容易适应你的需要。

这个呢?

import java.util.ArrayList;
import java.util.Collection;
import java.util.HashMap;

/**
  * @author ycoppel@google.com (Yohann Coppel)
  * 
  * @param <T>
  *          Object's type in the tree.
*/
public class Tree<T> {

  private T head;

  private ArrayList<Tree<T>> leafs = new ArrayList<Tree<T>>();

  private Tree<T> parent = null;

  private HashMap<T, Tree<T>> locate = new HashMap<T, Tree<T>>();

  public Tree(T head) {
    this.head = head;
    locate.put(head, this);
  }

  public void addLeaf(T root, T leaf) {
    if (locate.containsKey(root)) {
      locate.get(root).addLeaf(leaf);
    } else {
      addLeaf(root).addLeaf(leaf);
    }
  }

  public Tree<T> addLeaf(T leaf) {
    Tree<T> t = new Tree<T>(leaf);
    leafs.add(t);
    t.parent = this;
    t.locate = this.locate;
    locate.put(leaf, t);
    return t;
  }

  public Tree<T> setAsParent(T parentRoot) {
    Tree<T> t = new Tree<T>(parentRoot);
    t.leafs.add(this);
    this.parent = t;
    t.locate = this.locate;
    t.locate.put(head, this);
    t.locate.put(parentRoot, t);
    return t;
  }

  public T getHead() {
    return head;
  }

  public Tree<T> getTree(T element) {
    return locate.get(element);
  }

  public Tree<T> getParent() {
    return parent;
  }

  public Collection<T> getSuccessors(T root) {
    Collection<T> successors = new ArrayList<T>();
    Tree<T> tree = getTree(root);
    if (null != tree) {
      for (Tree<T> leaf : tree.leafs) {
        successors.add(leaf.head);
      }
    }
    return successors;
  }

  public Collection<Tree<T>> getSubTrees() {
    return leafs;
  }

  public static <T> Collection<T> getSuccessors(T of, Collection<Tree<T>> in) {
    for (Tree<T> tree : in) {
      if (tree.locate.containsKey(of)) {
        return tree.getSuccessors(of);
      }
    }
    return new ArrayList<T>();
  }

  @Override
  public String toString() {
    return printTree(0);
  }

  private static final int indent = 2;

  private String printTree(int increment) {
    String s = "";
    String inc = "";
    for (int i = 0; i < increment; ++i) {
      inc = inc + " ";
    }
    s = inc + head;
    for (Tree<T> child : leafs) {
      s += "\n" + child.printTree(increment + indent);
    }
    return s;
  }
}

如果您正在编写白板代码、进行面试，或者只是计划使用树，那么这些内容就有点冗长了。

应该进一步说，树不像Pair那样存在的原因是，你应该将你的数据封装在使用它的类中，最简单的实现是这样的:

/***
/* Within the class that's using a binary tree for any reason. You could 
/* generalize with generics IFF the parent class needs different value types.
 */
private class Node {
  public String value;
  public Node[] nodes; // Or an Iterable<Node> nodes;
}

这就是任意宽度的树。

如果你想要一个二叉树，它通常更容易使用命名字段:

private class Node { // Using package visibility is an option
  String value;
  Node left;
  Node right;
}

或者如果你想要一个trie

private class Node {
  String value;
  Map<char, Node> nodes;
}

现在你说你想要

给定一个表示给定节点的输入字符串，能够获得所有的子节点(某种类型的列表或字符串数组)

听起来像是你的家庭作业。 但既然我有理由相信任何最后期限都已经过去了……

import java.util.Arrays;
import java.util.ArrayList;
import java.util.List;

public class kidsOfMatchTheseDays {
 static private class Node {
   String value;
   Node[] nodes;
 }

 // Pre-order; you didn't specify.
 static public List<String> list(Node node, String find) {
   return list(node, find, new ArrayList<String>(), false);
 }

 static private ArrayList<String> list(
     Node node,
     String find,
     ArrayList<String> list,
     boolean add) {
   if (node == null) {
     return list;
   }
   if (node.value.equals(find)) {
     add = true;
   }
   if (add) {
     list.add(node.value);
   }
   if (node.nodes != null) {
     for (Node child: node.nodes) {
       list(child, find, list, add);
     }
   }
   return list;
 }

 public static final void main(String... args) {
   // Usually never have to do setup like this, so excuse the style
   // And it could be cleaner by adding a constructor like:
   //     Node(String val, Node... children) {
   //         value = val;
   //         nodes = children;
   //     }
   Node tree = new Node();
   tree.value = "root";
   Node[] n = {new Node(), new Node()};
   tree.nodes = n;
   tree.nodes[0].value = "leftish";
   tree.nodes[1].value = "rightish-leafy";
   Node[] nn = {new Node()};
   tree.nodes[0].nodes = nn;
   tree.nodes[0].nodes[0].value = "off-leftish-leaf";
   // Enough setup
   System.out.println(Arrays.toString(list(tree, args[0]).toArray()));
 }
}

这让你使用:

$ java kidsOfMatchTheseDays leftish
[leftish, off-leftish-leaf]
$ java kidsOfMatchTheseDays root
[root, leftish, off-leftish-leaf, rightish-leafy]
$ java kidsOfMatchTheseDays rightish-leafy
[rightish-leafy]
$ java kidsOfMatchTheseDays a
[]