在Java中是否有标准的Java库类来表示树?
具体来说,我需要表示以下内容:
任意节点上的子树可以有任意数量的子树 每个节点(根节点之后)及其子节点都有字符串值 我需要得到一个给定节点的所有子(某种类型的列表或字符串数组),它的字符串值(即。一个方法,将一个节点作为输入,并返回子节点的所有字符串值作为输出)
是否有任何可用的结构,或者我需要创建我自己的(如果是这样,实施建议将是伟大的)。
在Java中是否有标准的Java库类来表示树?
具体来说,我需要表示以下内容:
任意节点上的子树可以有任意数量的子树 每个节点(根节点之后)及其子节点都有字符串值 我需要得到一个给定节点的所有子(某种类型的列表或字符串数组),它的字符串值(即。一个方法,将一个节点作为输入,并返回子节点的所有字符串值作为输出)
是否有任何可用的结构,或者我需要创建我自己的(如果是这样,实施建议将是伟大的)。
当前回答
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
[]
其他回答
在这里:
public class Tree<T> {
private Node<T> root;
public Tree(T rootData) {
root = new Node<T>();
root.data = rootData;
root.children = new ArrayList<Node<T>>();
}
public static class Node<T> {
private T data;
private Node<T> parent;
private List<Node<T>> children;
}
}
这是一个基本的树结构,可用于String或任何其他对象。实现简单的树来满足您的需要是相当容易的。
您需要添加的只是用于添加、删除、遍历和构造函数的方法。节点是树的基本构建块。
请检查下面的代码,其中我使用了Tree数据结构,没有使用Collection类。代码可能有bug /改进,但请使用这只是作为参考
package com.datastructure.tree;
public class BinaryTreeWithoutRecursion <T> {
private TreeNode<T> root;
public BinaryTreeWithoutRecursion (){
root = null;
}
public void insert(T data){
root =insert(root, data);
}
public TreeNode<T> insert(TreeNode<T> node, T data ){
TreeNode<T> newNode = new TreeNode<>();
newNode.data = data;
newNode.right = newNode.left = null;
if(node==null){
node = newNode;
return node;
}
Queue<TreeNode<T>> queue = new Queue<TreeNode<T>>();
queue.enque(node);
while(!queue.isEmpty()){
TreeNode<T> temp= queue.deque();
if(temp.left!=null){
queue.enque(temp.left);
}else
{
temp.left = newNode;
queue =null;
return node;
}
if(temp.right!=null){
queue.enque(temp.right);
}else
{
temp.right = newNode;
queue =null;
return node;
}
}
queue=null;
return node;
}
public void inOrderPrint(TreeNode<T> root){
if(root!=null){
inOrderPrint(root.left);
System.out.println(root.data);
inOrderPrint(root.right);
}
}
public void postOrderPrint(TreeNode<T> root){
if(root!=null){
postOrderPrint(root.left);
postOrderPrint(root.right);
System.out.println(root.data);
}
}
public void preOrderPrint(){
preOrderPrint(root);
}
public void inOrderPrint(){
inOrderPrint(root);
}
public void postOrderPrint(){
inOrderPrint(root);
}
public void preOrderPrint(TreeNode<T> root){
if(root!=null){
System.out.println(root.data);
preOrderPrint(root.left);
preOrderPrint(root.right);
}
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
BinaryTreeWithoutRecursion <Integer> ls= new BinaryTreeWithoutRecursion <>();
ls.insert(1);
ls.insert(2);
ls.insert(3);
ls.insert(4);
ls.insert(5);
ls.insert(6);
ls.insert(7);
//ls.preOrderPrint();
ls.inOrderPrint();
//ls.postOrderPrint();
}
}
// TestTree.java
// A simple test to see how we can build a tree and populate it
//
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
import javax.swing.tree.*;
public class TestTree extends JFrame {
JTree tree;
DefaultTreeModel treeModel;
public TestTree( ) {
super("Tree Test Example");
setSize(400, 300);
setDefaultCloseOperation(EXIT_ON_CLOSE);
}
public void init( ) {
// Build up a bunch of TreeNodes. We use DefaultMutableTreeNode because the
// DefaultTreeModel can use it to build a complete tree.
DefaultMutableTreeNode root = new DefaultMutableTreeNode("Root");
DefaultMutableTreeNode subroot = new DefaultMutableTreeNode("SubRoot");
DefaultMutableTreeNode leaf1 = new DefaultMutableTreeNode("Leaf 1");
DefaultMutableTreeNode leaf2 = new DefaultMutableTreeNode("Leaf 2");
// Build our tree model starting at the root node, and then make a JTree out
// of it.
treeModel = new DefaultTreeModel(root);
tree = new JTree(treeModel);
// Build the tree up from the nodes we created.
treeModel.insertNodeInto(subroot, root, 0);
// Or, more succinctly:
subroot.add(leaf1);
root.add(leaf2);
// Display it.
getContentPane( ).add(tree, BorderLayout.CENTER);
}
public static void main(String args[]) {
TestTree tt = new TestTree( );
tt.init( );
tt.setVisible(true);
}
}
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;
}
}
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);
}
}
}