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
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;
}
}
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"));
}
}
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。