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
不使用Collection框架的Tree的自定义树实现。 它包含Tree实现所需的不同基本操作。
class Node {
int data;
Node left;
Node right;
public Node(int ddata, Node left, Node right) {
this.data = ddata;
this.left = null;
this.right = null;
}
public void displayNode(Node n) {
System.out.print(n.data + " ");
}
}
class BinaryTree {
Node root;
public BinaryTree() {
this.root = null;
}
public void insertLeft(int parent, int leftvalue ) {
Node n = find(root, parent);
Node leftchild = new Node(leftvalue, null, null);
n.left = leftchild;
}
public void insertRight(int parent, int rightvalue) {
Node n = find(root, parent);
Node rightchild = new Node(rightvalue, null, null);
n.right = rightchild;
}
public void insertRoot(int data) {
root = new Node(data, null, null);
}
public Node getRoot() {
return root;
}
public Node find(Node n, int key) {
Node result = null;
if (n == null)
return null;
if (n.data == key)
return n;
if (n.left != null)
result = find(n.left, key);
if (result == null)
result = find(n.right, key);
return result;
}
public int getheight(Node root){
if (root == null)
return 0;
return Math.max(getheight(root.left), getheight(root.right)) + 1;
}
public void printTree(Node n) {
if (n == null)
return;
printTree(n.left);
n.displayNode(n);
printTree(n.right);
}
}
其他回答
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。
不使用Collection框架的Tree的自定义树实现。 它包含Tree实现所需的不同基本操作。
class Node {
int data;
Node left;
Node right;
public Node(int ddata, Node left, Node right) {
this.data = ddata;
this.left = null;
this.right = null;
}
public void displayNode(Node n) {
System.out.print(n.data + " ");
}
}
class BinaryTree {
Node root;
public BinaryTree() {
this.root = null;
}
public void insertLeft(int parent, int leftvalue ) {
Node n = find(root, parent);
Node leftchild = new Node(leftvalue, null, null);
n.left = leftchild;
}
public void insertRight(int parent, int rightvalue) {
Node n = find(root, parent);
Node rightchild = new Node(rightvalue, null, null);
n.right = rightchild;
}
public void insertRoot(int data) {
root = new Node(data, null, null);
}
public Node getRoot() {
return root;
}
public Node find(Node n, int key) {
Node result = null;
if (n == null)
return null;
if (n.data == key)
return n;
if (n.left != null)
result = find(n.left, key);
if (result == null)
result = find(n.right, key);
return result;
}
public int getheight(Node root){
if (root == null)
return 0;
return Math.max(getheight(root.left), getheight(root.right)) + 1;
}
public void printTree(Node n) {
if (n == null)
return;
printTree(n.left);
n.displayNode(n);
printTree(n.right);
}
}
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
[]
我对所有这些方法都有意见。
我使用的是“MappedTreeStructure”实现。这个实现很好地重新组织了树,并且不包含节点的“副本”。
但是没有提供分级方法。
看看那些有问题的输出!
MutableTree<String> tree = new MappedTreeStructure<>();
tree.add("0", "1");
tree.add("0", "2");
tree.add("0", "3");
tree.add("0", "4");
tree.add("0", "5");
tree.add("2", "3");
tree.add("2", "5");
tree.add("1", "2");
tree.add("1", "3");
tree.add("1", "5");
System.out.println(
tree.toString()
);
哪个输出:(错误)
- 0
- 1
- 2
- 3
- 5
- 4
还有这个:(正确)
tree = new MappedTreeStructure<>();
tree.add("0", "1");
tree.add("0", "2");
tree.add("0", "3");
tree.add("0", "4");
tree.add("0", "5");
tree.add("1", "2");
tree.add("1", "3");
tree.add("1", "5");
tree.add("2", "3");
tree.add("2", "5");
System.out.println(
tree.toString()
);
正确的输出:
- 0
- 1
- 2
- 3
- 5
- 4
如此!我创建了另一个实现来欣赏。请给一些建议和反馈!
package util;
import java.util.HashMap;
import java.util.Map;
public class Node<N extends Comparable<N>> {
public final Map<N, Node<N>> parents = new HashMap<>();
public final N value;
public final Map<N, Node<N>> children = new HashMap<>();
public Node(N value) {
this.value = value;
}
}
package util;
import java.util.*;
import java.util.stream.Collectors;
public class HierarchyTree<N extends Comparable<N>> {
protected final Map<N, Node<N>> nodeList = new HashMap<>();
public static <T extends Comparable<T>> Node<T> state(Map<T, Node<T>> nodeList, T node) {
Node<T> tmp = nodeList.getOrDefault(node, new Node<>(node));
nodeList.putIfAbsent(node, tmp);
return tmp;
}
public static <T extends Comparable<T>> Node<T> state(Map<T, Node<T>> nodeList, Node<T> node) {
Node<T> tmp = nodeList.getOrDefault(node.value, node);
nodeList.putIfAbsent(node.value, tmp);
return tmp;
}
public Node<N> state(N child) {
return state(nodeList, child);
}
public Node<N> stateChild(N parent, N child) {
Node<N> pai = state(parent);
Node<N> filho = state(child);
state(pai.children, filho);
state(filho.parents, pai);
return filho;
}
public List<Node<N>> addChildren(List<N> children) {
List<Node<N>> retorno = new LinkedList<>();
for (N child : children) {
retorno.add(state(child));
}
return retorno;
}
public List<Node<N>> addChildren(N parent, List<N> children) {
List<Node<N>> retorno = new LinkedList<>();
for (N child : children) {
retorno.add(stateChild(parent, child));
}
return retorno;
}
public List<Node<N>> addChildren(N parent, N... children) {
return addChildren(parent, Arrays.asList(children));
}
public List<Node<N>> getRoots() {
return nodeList.values().stream().filter(value -> value.parents.size() == 0).collect(Collectors.toList());
}
@Override
public String toString() {
return deepPrint("- ");
}
public String deepPrint(String prefix) {
StringBuilder builder = new StringBuilder();
deepPrint(builder, prefix, "", getRoots());
return builder.toString();
}
protected void deepPrint(StringBuilder builder, String prefix, String sep, List<Node<N>> node) {
for (Node<N> item : node) {
builder.append(sep).append(item.value).append("\n");
deepPrint(builder, prefix, sep + prefix, new ArrayList<>(item.children.values()));
}
}
public SortedMap<Long, Set<N>> tree() {
SortedMap<Long, Set<N>> tree = new TreeMap<>();
tree(0L, tree, getRoots());
return tree;
}
protected void tree(Long i, SortedMap<Long, Set<N>> tree, List<Node<N>> roots) {
for (Node<N> node : roots) {
Set<N> tmp = tree.getOrDefault(i, new HashSet<>());
tree.putIfAbsent(i, tmp);
tmp.add(node.value);
tree(i + 1L, tree, new ArrayList<>(node.children.values()));
}
}
public void prune() {
Set<N> nodes = new HashSet<>();
SortedMap<Long, Set<N>> tree = tree();
List<Long> treeInverse = tree.keySet().stream().sorted(Comparator.reverseOrder()).collect(Collectors.toList());
for (Long treeItem : treeInverse) {
for (N n : tree.get(treeItem)) {
Map<N, Node<N>> children = nodeList.get(n).children;
for (N node : nodes) {
children.remove(node);
}
nodes.addAll(children.keySet());
}
}
}
public static void main(String[] args) {
HierarchyTree<Integer> tree = new HierarchyTree<>();
tree.addChildren(Arrays.asList(1, 2, 3, 4, 5));
tree.addChildren(1, Arrays.asList(2, 3, 5));
tree.addChildren(2, Arrays.asList(3, 5));
tree.prune();
System.out.println(tree);
tree = new HierarchyTree<>();
tree.addChildren(Arrays.asList(1, 2, 3, 4, 5));
tree.addChildren(2, Arrays.asList(3, 5));
tree.addChildren(1, Arrays.asList(2, 3, 5));
tree.prune();
System.out.println(tree);
}
}
输出总是正确的:
1
- 2
- - 3
- - 5
4
1
- 2
- - 3
- - 5
4
我编写了一个树库,它可以很好地使用Java8,并且没有其他依赖项。它还提供了对函数式编程的一些思想的松散解释,并允许您映射/过滤/修剪/搜索整个树或子树。
https://github.com/RutledgePaulV/prune
这个实现在索引方面没有做任何特别的事情,而且我也没有偏离递归,所以使用大型树的性能可能会下降,可能会破坏堆栈。但如果你所需要的只是一个简单的小到中等深度的树,我认为它已经足够好了。它提供了一个健全的(基于值的)相等定义,它还有一个toString实现,可以让您可视化树!