Question

运用Java数据结构的知识创建树，内容是族谱。

Answer 1

每一个节点有一个成员变量引用下一个节点就行了。

大致实现了一下单向链表 没有加入异常也没有仔细考虑实现的代码的效率，可以参考下。

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public class LinkListTest {
    public static void main(String[] args) {
        LinkList<String> ll=new LinkList<String>();
        ll.add("a");
        ll.add("b");
        ll.add("c");
        ll.add("d");
        ll.remove(1);
        System.out.println(ll.get(0));
        System.out.println(ll.get(1));
        System.out.println(ll.get(2));
        System.out.println(ll.get(3));
        System.out.println(ll.size());
    }
}
class LinkList<T>{
    private Node<T> frist=null;
    private Node<T> last=null;
    private int size=0;
                                                                          
    public void add(T t){
        if(frist==null){
            Node<T> node=new Node<T>();
            node.setT(t);
            size++;
            frist=node;
            last=node;
        }else{        
            Node<T> node=new Node<T>();       
            node.setT(t);
            last.setNextNode(node);
            size++;
            last=node;
        }
    }
                                                                          
    public T get(int i){      
        if(i>=0&&i<size){
            Node<T> nod=null;
            for(int n=0;n<=i;n++){
                if(n==0)
                   nod=frist;
                else
                   nod=nod.getNextNode();             
                if(i==n){
                    return nod.getT();
                }         
            }         
        }
         return null;
    } 
                                                                          
    public void remove(int i){
        if(i>=0&&i<size){
           if(size<2){
               frist=null;
               last=null;
               size=0;
           }else{
               size--;
              if(i==0){
                 frist=frist.getNextNode();        
              }else{      
                    Node<T> nod1=null;
                    Node<T> nod2=null;
                    for(int n=0;n<=i;n++){
                        if(n==0){
                           nod1=frist;
                           nod2=frist;
                        }else{
                           nod2=nod1;
                           nod1=nod1.getNextNode();                      
                        } 
                        if(i==n){
                            if(nod1!=null)
                                nod2.setNextNode(nod1.getNextNode());
                            else{
                                nod2.setNextNode(null);
                                last=nod2;
                            }
                        }             
                    } 
              }   
            }
        }
    }
                                                                          
                                                                          
                                                                          
    public int size(){
        return size;
    }
}
class Node <T>{
    public T getT() {
        return t;
    }
    public void setT(T t) {
        this.t = t;
    }
    public Node<T> getNextNode() {
        return nextNode;
    }
    public void setNextNode(Node<T> nextNode) {
        this.nextNode = nextNode;
    }
    private T t;
    private Node<T> nextNode=null;
}

Answer 2

数据源是在数据库里面查出来的么

追问

嗯嗯 是的 求帮助

追答

       1

   2       2

3     3      3 

先查出最高层级，生成一个树，然后在展开的时候把这个id穿到后台，查处当前这个树的子节点，如此一成一层弄就好了，例子如图