由于问题要求可用的数据结构，树可以由列表或数组构造:

Object[] tree = new Object[2];
tree[0] = "Hello";
{
  Object[] subtree = new Object[2];
  subtree[0] = "Goodbye";
  subtree[1] = "";
  tree[1] = subtree;
}

Instanceof可用于确定元素是子树还是终端节点。

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?

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
[]

Java中有一些树数据结构，比如JDK Swing中的DefaultMutableTreeNode, Stanford解析器包中的tree，以及其他一些玩具代码。但这些都不够，也不够小，不能用于一般用途。

Java tree项目试图在Java中提供另一种通用的树数据结构。这个和其他的区别是

Totally free. You can use it anywhere (except in your homework :P) Small but general enough. I put everything of the data structure in one class file, so it would be easy to copy/paste. Not just a toys. I am aware of dozens of Java tree codes that can only handle binary trees or limited operations. This TreeNode is much more than that. It provides different ways of visiting nodes, such as preorder, postorder, breadthfirst, leaves, path to root, etc. Moreover, iterators are provided too for the sufficiency. More utils will be added. I am willing to add more operations to make this project comprehensive, especially if you send a request through github.

在过去，我只是为此使用了一个嵌套映射。这是我今天用的，它很简单，但它符合我的需要。也许这能帮到另一个人。

import com.fasterxml.jackson.annotation.JsonValue;
import com.fasterxml.jackson.databind.ObjectMapper;

import java.util.HashMap;
import java.util.Map;
import java.util.TreeMap;

/**
 * Created by kic on 16.07.15.
 */
public class NestedMap<K, V> {
    private final Map root = new HashMap<>();

    public NestedMap<K, V> put(K key) {
        Object nested = root.get(key);

        if (nested == null || !(nested instanceof NestedMap)) root.put(key, nested = new NestedMap<>());
        return (NestedMap<K, V>) nested;
    }

    public Map.Entry<K,V > put(K key, V value) {
        root.put(key, value);

        return (Map.Entry<K, V>) root.entrySet().stream().filter(e -> ((Map.Entry) e).getKey().equals(key)).findFirst().get();
    }

    public NestedMap<K, V> get(K key) {
        return (NestedMap<K, V>) root.get(key);
    }

    public V getValue(K key) {
        return (V) root.get(key);
    }

    @JsonValue
    public Map getRoot() {
        return root;
    }

    public static void main(String[] args) throws Exception {
        NestedMap<String, Integer> test = new NestedMap<>();
        test.put("a").put("b").put("c", 12);
        Map.Entry<String, Integer> foo = test.put("a").put("b").put("d", 12);
        test.put("b", 14);
        ObjectMapper mapper = new ObjectMapper();
        System.out.println(mapper.writeValueAsString(test));

        foo.setValue(99);
        System.out.println(mapper.writeValueAsString(test));

        System.out.println(test.get("a").get("b").getValue("d"));
    }
}