二进制搜索树打印范围

Question

因此，我必须修改BST类以包含一个PrintRange函数，该函数实际上将按顺序打印两个值之间的所有节点。

这是课程

/** Source code example for "A Practical Introduction to Data
    Structures and Algorithm Analysis, 3rd Edition (Java)" 
    by Clifford A. Shaffer
    Copyright 2008-2011 by Clifford A. Shaffer
*/

import java.lang.Comparable;

/** Binary Search Tree implementation for Dictionary ADT */
class BST<Key extends Comparable<? super Key>, E>
         implements Dictionary<Key, E> {
  private BSTNode<Key,E> root; // Root of the BST
  int nodecount;             // Number of nodes in the BST

  /** Constructor */
  BST() { root = null; nodecount = 0; }

  /** Reinitialize tree */
  public void clear() { root = null; nodecount = 0; }

  /** Insert a record into the tree.
      @param k Key value of the record.
      @param e The record to insert. */
  public void insert(Key k, E e) {
    root = inserthelp(root, k, e);
    nodecount++;
  }

// Return root

  public BSTNode getRoot()
  {
   return root;
  }  

 /** Remove a record from the tree.
      @param k Key value of record to remove.
      @return The record removed, null if there is none. */

  public E remove(Key k) {
    E temp = findhelp(root, k);   // First find it
    if (temp != null) {
      root = removehelp(root, k); // Now remove it
      nodecount--;
    }
    return temp;
  }

  /** Remove and return the root node from the dictionary.
      @return The record removed, null if tree is empty. */
  public E removeAny() {
    if (root == null) return null;
    E temp = root.element();
    root = removehelp(root, root.key());
    nodecount--;
    return temp;
  }

  /** @return Record with key value k, null if none exist.
      @param k The key value to find. */
  public E find(Key k) { return findhelp(root, k); }

  /** @return The number of records in the dictionary. */
  public int size() { return nodecount; }

  private E findhelp(BSTNode<Key,E> rt, Key k) {
  if (rt == null) return null;
  if (rt.key().compareTo(k) > 0)
    return findhelp(rt.left(), k);
  else if (rt.key().compareTo(k) == 0) return rt.element();
  else return findhelp(rt.right(), k);
}
/** @return The current subtree, modified to contain
   the new item */
private BSTNode<Key,E> inserthelp(BSTNode<Key,E> rt,
                                  Key k, E e) {
  if (rt == null) return new BSTNode<Key,E>(k, e);
  if (rt.key().compareTo(k) > 0)
    rt.setLeft(inserthelp(rt.left(), k, e));
  else
    rt.setRight(inserthelp(rt.right(), k, e));
  return rt;
}
/** Remove a node with key value k
    @return The tree with the node removed */

 private BSTNode<Key,E> removehelp(BSTNode<Key,E> rt,Key k) {
  if (rt == null) return null;
  if (rt.key().compareTo(k) > 0)
    rt.setLeft(removehelp(rt.left(), k));
  else if (rt.key().compareTo(k) < 0)
    rt.setRight(removehelp(rt.right(), k));
  else { // Found it
    if (rt.left() == null) return rt.right();
    else if (rt.right() == null) return rt.left();
    else { // Two children
      BSTNode<Key,E> temp = getmin(rt.right());
      rt.setElement(temp.element());
      rt.setKey(temp.key());
      rt.setRight(deletemin(rt.right()));
    }
  }
  return rt;
}

private BSTNode<Key,E> getmin(BSTNode<Key,E> rt) {
  if (rt.left() == null) return rt;
  return getmin(rt.left());
}

private BSTNode<Key,E> deletemin(BSTNode<Key,E> rt) {
  if (rt.left() == null) return rt.right();
  rt.setLeft(deletemin(rt.left()));
  return rt;
}
  private void printhelp(BSTNode<Key,E> rt) {
    if (rt == null) return;
    printhelp(rt.left());
    printVisit(rt.element());
    printhelp(rt.right());
  }

  private StringBuffer out;

  public String toString() {
    out = new StringBuffer(400);
    printhelp(root);
    return out.toString();
  }
  private void printVisit(E it) {
    out.append(it + "\n");
  }

  public void printPreOrder(BSTNode<E, E> root) {
      if (root != null) {
          System.out.println(root.element());
          printPreOrder(root.left());
          printPreOrder(root.right());
      }
  }

  public void printInOrder(BSTNode<E, E> root) {
      if (root != null) {
          printInOrder(root.left());
          System.out.println(root.element());
          printInOrder(root.right());
      }
  }

  public void printPostOrder(BSTNode<E, E> root) {
      if (root != null) {
          printPostOrder(root.left());
          printPostOrder(root.right());
          System.out.println(root.element());
      }
  }

}

到目前为止，这是我对PrintRange函数的了解：

public void printRange(BSTNode<E, E> root, E low, E high) {
          if (root != null) {
            printRange(root.left(), low, high);
            if (root.element().toString().compareTo(low.toString()) > 0 && root.element().toString().compareTo(high.toString()) < 0)
              System.out.println(root.element());
            printRange(root.right(), low, high);
          }
      }

但这给我一个错误。 关于如何比较元素/节点/我什至在BST中不确定的任何建议？

这是司机，如果有帮助

import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;

public class Lab8a {

    public static void main(String[] args) {
        BST<String, String> tree = new BST<String, String>();
        Scanner fileScan = null, scan = new Scanner(System.in);

        //Open file
        try {
            fileScan = new Scanner(new File("inventory.txt"));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }

        //Reads elements from file
        while (fileScan.hasNextLine()) {
            String s = fileScan.nextLine();
            tree.insert(s, s);
        }

        System.out.println("\nRange");
        tree.printRange(tree.getRoot(), "A", "B");

    }

}

和文本文件：

CT16C1288B

DT14B1225F

MI15B1250A

MI15B1251A

HO03N1095A

HY07D1095BQ

KI04D2593C

DG12A1240AQ

HY03G2593BQ

TO30A1310A

HO03N1095AQ

HO01H1351C

HO01H1350C

FT18A1288B

LR15A1000A

BM12E1000A

VW02B3113A

NI23H1230AQ

LX03D2503A

LX03D2502A

LX03D2502A

VW22A3113B

VW22B3113A

Answer 1

我误解了。 没有错误。 我必须在某个时候修复代码。