如何在我的树 java 中使用另一个 'helper' 类创建 toString？

Question

我正试图找出如何从下面给出的parentheticRepresentation类创建toString()方法。

public static <E> String parentheticRepresentation(Tree<E> T, Position<E> v) {
        String s = v.element().toString();
        if (T.islnternal(v)) {
        Boolean firstTime = true;
        for (Position<E> w : T.children(v))
        if (firstTime) {
        s += " ( " + parentheticRepresentation(T, w);
        firstTime = false;
        }
        else s += ", " + parentheticRepresentation(T, w);
        s += " ) ";
        }
        return s;
        }

在我的主类中，当我为树创建节点并尝试输出整个树时，它只输出一个带有parentheticrepresentation的节点。那么我如何使用这个来创建另一个toString()类，以便当我调用输出树时，它给我上面类中的表示。任何帮助将不胜感激!

public static void main(String[] args) {
    LinkedTree<Character> T = new LinkedTree();
    // add root
    T.addRoot('A');
    // add children of root
    T.createNode('B', (TreeNode) (T.root()), new NodePositionList());
    TreePosition C = T.createNode('C', (TreeNode) (T.root()),
            new NodePositionList());
    T.createNode('D', (TreeNode) (T.root()), new NodePositionList());
    // add children of node C
    T.createNode('E', C, new NodePositionList());
    TreePosition F = T.createNode('F', C, new NodePositionList());
    T.createNode('G', C, new NodePositionList());
    // add childrn of Node F
    T.createNode('H', F, new NodePositionList());
    T.createNode('I', F, new NodePositionList());
    // print out tree
    System.out.println("Size = " + T.size());
    System.out.println("Here is the tree:");
    System.out.println(T);
}

}

Answer 1

您需要做一个预先排序遍历，只是在遍历子元素时打开括号，然后关闭它。

类似这样的东西(不确定它是否有效，因为我没有测试它，但只是给你一个想法)

public static <E> String parentheticRepresentation(Tree<E> T, Position<E> v) {
    String s = "";
    if (T.islnternal(v)) {
        s = v.element().toString(); //adds node to string      
        if(T.children(v).size() > 0)
        {
            s += "("; //opens parenthesis for children
            for (Position<E> w : T.children(v))
            {
                s += parentheticRepresentation(T, w) + ",";
            }
            s = s.deleteCharAt(s.length() - 1); //deletes last comma
            s += ")"; //closes parenthesis for children
        }
    }
    return s;
}

要重写toString()方法，请在Tree类中添加如下内容:

@Override
public String toString(){
     //Just speculating the TreeHelper name, but its calling the helper method.
     return TreeHelper.parentheticRepresentation(this,this.root());
}

编辑:使用"this"引用，因为我们在Tree类中。

Answer 2

很明显，OP在我的数据结构类....

这是我所做的(使用给我们的括号表示方法)。

private static <E> String parentheticRepresentation (Tree <E> T, Position <E> v){
    String s = v.element().toString();
    if (T.isInternal(v)){
        Boolean firstTime = true;

        Iterator <Position<E>> it = T.children(v).iterator();
        while (it.hasNext()){
            Position<E> w = it.next();

            if (firstTime) {
                s+= " ( " + parentheticRepresentation (T,w);
                firstTime = false;
            }
            else s+= ", " + parentheticRepresentation (T,w);
        s+= " )";
    }}
    return s;
            }

public String toString() 
{
return parentheticRepresentation(this,this.root());
}

问题是，我不断得到堆栈溢出错误为我的子方法(下面添加)

编辑:我修复了堆栈溢出错误，但知道我有与op相同的问题。只有根(A)输出。

public Position<E> parent(Position<E> v) throws InvalidPositionException,
        BoundaryViolationException {
    TreePosition<E> p = checkPosition(v);
    Position<E> parentPosition = p.getParent();
    if (parentPosition == null)
        throw new BoundaryViolationException("No parent");
    return parentPosition;
}
public Iterable<Position<E>> children(Position<E> v)
        throws InvalidPositionException {
    TreePosition <E> p = checkPosition(v);
    if (isExternal(v))    //STACK OVERFLOW
        throw new InvalidPositionException("");
return p.getChildren();
}
public boolean isInternal(Position<E> v) throws InvalidPositionException {
    checkPosition(v);
    return (children(v) != null);
}
public boolean isExternal(Position<E> v) throws InvalidPositionException {
    checkPosition(v);  
    return (children(v) == null); // STACK OVERFLOW
}