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!