二进制搜索树toString [英] Binary Search Tree toString

问题描述

我无法以教授想要的格式打印二进制搜索树.

I am having trouble printing out a binary search tree in the format my professor wants.

他的格式如下:

{(12,10,13),(10,8,11),(8,6,9),(6,4,7),(4,2,5),(2,1,3),(1,*,*),(3,*,*),(5,*,*),(7,*,*),(9,*,*),(11,*,*),(13,*,*)} 

子集中的第一个数字是根节点，左边和右边的节点是左边和右边的子节点.然后，循环迭代后，左子节点将成为根节点.一切正常，直到我到达子集中只有一个节点的位置.它只会打印(1，*，*)到最后，而我不知道该怎么做.是否可以递归地执行此toString方法?

The first number in a subset is the root node, and the left and right nodes are the left and right children. The left child then becomes the root node after a loop iterates. everything works fine until I reach where only one node exists in a subset. It just prints (1, *, *) till the end, and I can not figure out how to do it another way. Is it possible to recursively do this toString method?

我的代码:

public String toString()
{
   if (root == null)
       return "{}";
   String str = "{";
   Node tmp = root;
   for (int i = 0; i < size; i++)
   {   
        if (tmp.right != null && tmp.left == null)
            str += "("+tmp.data+", "+tmp.right.data+", *)";
        if (tmp.left != null && tmp.right == null)
            str += "("+tmp.data+", "+tmp.left.data+", *)";
        if (tmp.left == null && tmp.right == null)          
            str += "("+tmp.data+", *, *)";
        else
            str += "("+tmp.data+", "+tmp.left.data+", "+tmp.right.data+")";

        if (tmp.left != null)
            tmp = tmp.left;
   }

   return str += "}";   
}

推荐答案

此方法取决于您如何设置对象，但是我通常有一个执行递归操作的Node类.如果以这种方式实施，您应该会看到类似

This approach depends how you have your objects setup, but I typically have a Node class that performs the recursive operations. If implemented in this fashion, you should see output like so

{(12,10,13),(10,8,11),(8,*,*),(11,*,*),(13,*,*)}

在此示例中，我们将有一个方法返回Node类上的(data,left,right)格式.

For this example, we will have a method that returns your (data,left,right) format on the Node class.

public class Node<T>

    protected T data;
    protected Node<T> left;
    protected Node<T> right;

    public String tuple() {
        StringBuilder sb = new StringBuilder("(")
                .append(this.data)
                .append(",");
        sb.append(this.left == null ? "*" : this.left.data)
                .append(",");
        sb.append(this.right == null ? "*" : this.right.data)
                .append(")");
        return sb.toString();
    }

    // other methods
}

然后，递归字符串将像这样在toString中实现.

Then, the recursive string would be implemented in the toString like so.

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        String divider = ",";

        sb.append(this.tuple()).append(divider);

        if (this.left != null) {
            sb.append(this.left).append(divider); // recurse left
        }

        if (this.right != null) {
            sb.append(this.right).append(divider); // recurse right
        }

        if (sb.length() > divider.length() - 1) {
            sb.setLength(sb.length() - divider.length());
        }

        return sb.toString();
    }
}

然后，在某些BinaryTree类中，您可以拥有

Then, in some BinaryTree class, you could have

public class BinaryTree<E extends Comparable<? super E>> {

    protected Node<E> root;

    @Override
    public String toString() {
        return "{"
                + (root == null ? "" : String.valueOf(this.root)) +
                "}";
    }

    // other methods

}

这篇关于二进制搜索树toString的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！