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社区首页 >问答首页 >多父母多子图的实现

多父母多子图的实现
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Stack Overflow用户
提问于 2020-01-25 03:54:32
回答 1查看 3.8K关注 0票数 3

我需要实现一个类似树的数据结构,其中每个节点都有多个父节点和子节点,因此有多个根。

新节点将仅作为没有父节点的根或一个或多个父节点的子节点添加到树中。

节点不会从子节点开始,但是任何现有节点都可以获得具有任意数量的其他父节点的另一个子节点。

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回答 1

Stack Overflow用户

回答已采纳

发布于 2020-01-26 18:02:18

我能够创建一个Node类和Graph类。我给每个Node一个父节点和子节点的List<Node>

每个Node都跟踪自己的直接子女和父母。

节点实现

代码语言:javascript
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public class Node<T>{
    private T data;
    private List<Node<T>> parents;
    private List<Node<T>> children = new ArrayList<>();//this can be initialized because a node will not start with children
    public Node(T data){//adding a node without parents
        this.data = data;
        parents = new ArrayList<>();//make parents an empty ArrayList so other methods won't break with a null value
    }
    public Node(T data, List<Node<T>> parents){//adding a node with parents
        this.data = data;
        this.parents = parents;
    }

    //search methods
    public List<Node<T>> getChildren(){return children;}//return only direct children
    public List<Node<T>> getChildren(int level){return getChildren(new ArrayList<>(Collections.singletonList(this)),new ArrayList<>(),level);}//convenience method to find only this node's children to a certain level
    public List<Node<T>> getChildren(List<Node<T>> find, List<Node<T>> found, int level){//level can be -1 to search through all children or a positive integer to only search the first n level of children.
        if(level!=0){//setting level to -1 will never stop the search until all children have been found because level only goes down, because the level will never reach zero
            for(Node<T> node:find){//can find the children of multiple nodes
                if(node.hasChild()) {
                    for (Node<T> child : node.getChildren()) {
                        if (!found.contains(child)) {//trees can intersect, so the child may have already been found, so only add if it hasn't
                            found.add(child);
                        }
                    }
                    getChildren(node.getChildren(),found,level--);//recursively find the remaining children of the current node
                }
            }
        }
        return found;
    }
    //a method that finds parents can be implemented by modifying the getChildren() methods

    //examples of other methods that can be added
    public T getData(){return data;}
    public boolean hasChild(){return children.size()>0;}
    void addChild(Node<T> node){children.add(node);}
    void addChildren(List<Node<T>> nodes){children.addAll(nodes);}
    public List<Node<T>> getParents(){return parents;}
    public boolean hasParent(){return parents.size()>0;}
    void addParent(Node<T> node){parents.add(node);}
    void addParent(List<Node<T>> nodes){parents.addAll(nodes);}
}

Graph类仅用于跟踪图的根。根只是一个没有父级的Node。每个方法的范围可以根据您的特定用例进行调整。这个类是由一个更特定的类来扩展的,这个类包括处理特定数据类型的方法。

图的实现:

代码语言:javascript
运行
复制
public class Graph<T> {
    private List<Node<T>> roots;
    protected Tree(List<Node<T>> roots){this.roots = roots;}//Graph class can be initialized with or without existing roots 
    protected Tree(){roots = new ArrayList<>();}
    public List<Node<T>> getRoots(){return roots;}
    public List<Node<T>> getAllNodes(){
        List<Node<T>> nodes = roots.get(0).getChildren(roots,new ArrayList<>(),-1);//loop through all roots with an empty list for nodes already found, because no nodes have been found yet
        nodes.addAll(roots);
        return nodes;
    }
    public void addNode(Node<T> node){
        for(Node<T> parent:node.getParents()){//for each parent node add this node as their child
            parent.addChild(node);
        }
        if(!node.hasParent())roots.add(node);
    }
    public void addNodes(List<Node<T>> nodes){
        for(Node<T> node:nodes){
            addNode(node);
        }
    }
}

当添加带有父节点的新Node时,Graph类将获取父节点并将Node添加为这些父节点的子节点,以便父节点知道它们有子节点。

票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/59906194

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