二叉搜索树的字符串表示

我一直在尝试为二叉搜索树编写一个递归字符串方法，该方法返回具有预订路径信息的树的多行表示 .

每个节点都应该以一系列<和>字符开头，这些字符显示从根到该节点的路径 . 我不确定如何使用每个连续调用一个字符扩展的字符串前缀参数 .

该方法应该能够重现这个例子：

树：

15
    /  \
   12  18
  /   /  \
 10  16  20
   \   \
   11  17

预期打印输出：

15
<12
<<10
<<>11
>18
><16
><>17
>>20

我是递归的新手，到目前为止，我的实际打印输出在经过几个小时的代码处理后还不够接近：

18
<17
<10
>15
<11
>12
16
20

这是我的树节点类正常工作：

/**
 * A single binary tree node.
 * <p>
 * Each node has both a left or right child, which can be null.
 */
public class TreeNode<E> {

  private E data;
  private TreeNode<E> left;
  private TreeNode<E> right;

  /**
   * Constructs a new node with the given data and references to the
   * given left and right nodes.
   */
  public TreeNode(E data, TreeNode<E> left, TreeNode<E> right) {
    this.data = data;
    this.left = left;
    this.right = right;
  }

  /**
   * Constructs a new node containing the given data.
   * Its left and right references will be set to null.
   */
  public TreeNode(E data) {
    this(data, null, null);
  }

  /** Returns the item currently stored in this node. */
  public E getData() {
    return data;
  }

  /** Overwrites the item stored in this Node with the given data item. */
  public void setData(E data) {
    this.data = data;
  }

  /**
   * Returns this Node's left child.
   * If there is no left left, returns null.
   */
  public TreeNode<E> getLeft() {
    return left;
  }

  /** Causes this Node to point to the given left child Node. */
  public void setLeft(TreeNode<E> left) {
    this.left = left;
  }

  /**
   * Returns this nodes right child.
   * If there is no right child, returns null.
   */
  public TreeNode<E> getRight() {
    return right;
  }

  /** Causes this Node to point to the given right child Node. */
  public void setRight(TreeNode<E> right) {
    this.right = right;
  }
}

这是我的二叉搜索树类，底部附近有toFullString（）方法：

import java.util.*;

/**
 * A binary search tree (BST) is a sorted ADT that uses a binary
 * tree to keep all elements in sorted order.  If the tree is
 * balanced, performance is very good: O(n lg n) for most operations.
 * If unbalanced, it performs more like a linked list: O(n).
 */
public class BinarySearchTree<E extends Comparable<E>> {

  private TreeNode<E> root = null;
  private int size = 0;

  /** Creates an empty tree. */
  public BinarySearchTree() {
  }


    public BinarySearchTree(Collection<E> col) {
        List<E> list = new ArrayList<E>(col);
        Collections.shuffle(list);
        for (int i = 0; i < list.size() ; i++) {
            add(list.get(i));
        }
    }


  /** Adds the given item to this BST. */
  public void add(E item) {
    this.size++;
    if (this.root == null) {
      //tree is empty, so just add item
      this.root = new TreeNode<E>(item);
    }else {
      //find where to insert, with pointer to parent node
      TreeNode<E> parent = null;
      TreeNode<E> curr = this.root;
      boolean wentLeft = true;
      while (curr != null) {  //will execute at least once
        parent = curr;
        if (item.compareTo(curr.getData()) <= 0) {
          curr = curr.getLeft();
          wentLeft = true;
        }else {
          curr = curr.getRight();
          wentLeft = false;
        }
      }

      //now add new node on appropriate side of parent
      curr = new TreeNode<E>(item);
      if (wentLeft) {
        parent.setLeft(curr);
      }else {
        parent.setRight(curr);
      }
    }
  }

  /** Returns the greatest (earliest right-most node) of the given tree. */
  private E findMax(TreeNode<E> n) {
    if (n == null) {
      return null;
    }else if (n.getRight() == null) {
      //can't go right any more, so this is max value
      return n.getData();
    }else {
      return findMax(n.getRight());
    }
  }

  /**
   * Returns item from tree that is equivalent (according to compareTo)
   * to the given item.  If item is not in tree, returns null.
   */
  public E get(E item) {
    return get(item, this.root);
  }

  /** Finds it in the subtree rooted at the given node. */
  private E get(E item, TreeNode<E> node) {
    if (node == null) {
      return null;
    }else if (item.compareTo(node.getData()) < 0) {
      return get(item, node.getLeft());
    }else if (item.compareTo(node.getData()) > 0) {
      return get(item, node.getRight());
    }else {
      //found it!
      return node.getData();
    }
  }

  /**
   * Removes the first equivalent item found in the tree.
   * If item does not exist to be removed, throws IllegalArgumentException().
   */
  public void remove(E item) {
    this.root = remove(item, this.root);
  }

  private TreeNode<E> remove(E item, TreeNode<E> node) {
    if (node == null) {
      //didn't find item
      throw new IllegalArgumentException(item + " not found in tree.");
    }else if (item.compareTo(node.getData()) < 0) {
      //go to left, saving resulting changes made to left tree
      node.setLeft(remove(item, node.getLeft()));
      return node;
    }else if (item.compareTo(node.getData()) > 0) {
      //go to right, saving any resulting changes
      node.setRight(remove(item, node.getRight()));
      return node;
    }else {
      //found node to be removed!
      if (node.getLeft() == null && node.getRight() == null) {
        //leaf node
        return null;
      }else if (node.getRight() == null) {
        //has only a left child
        return node.getLeft();
      }else if (node.getLeft() == null) {
        //has only a right child
        return node.getRight();
      }else {
        //two children, so replace the contents of this node with max of left tree
        E max = findMax(node.getLeft());  //get max value
        node.setLeft(remove(max, node.getLeft())); //and remove its node from tree
        node.setData(max);
        return node;
      }
    }
  }

  /** Returns the number of elements currently in this BST. */
  public int size() {
    return this.size;
  }

  /**
   * Returns a single-line representation of this BST's contents.
   * Specifically, this is a comma-separated list of all elements in their
   * natural Comparable ordering.  The list is surrounded by [] characters.
   */
  @Override
  public String toString() {
    return "[" + toString(this.root) + "]";
  }

  private String toString(TreeNode<E> n) {
    //would have been simpler to use Iterator... but not implemented yet.
    if (n == null) {
      return "";
    }else {
      String str = "";
      str += toString(n.getLeft());
      if (!str.isEmpty()) {
        str += ", ";
      }
      str += n.getData();
      if (n.getRight() != null) {
        str += ", ";
        str += toString(n.getRight());
      }
      return str;
    }
  }

    public String toFullString() {
        StringBuilder sb = new StringBuilder();
        toFullString(root, sb);
        return sb.toString();
    }

  /**
   * Preorder traversal of the tree that builds a string representation
   * in the given StringBuilder.
   * @param n root of subtree to be traversed
   * @param sb StringBuilder in which to create a string representation
   */
  private void toFullString(TreeNode<E> n, StringBuilder sb)
  {

    if (n == null)
    {
      return;
    }



sb.append(n.getData().toString());
    sb.append("\n");


        if (n.getLeft() != null) {
        sb.append("<");
    } else if (n.getRight() != null) {
        sb.append(">");
    }

    if (n.getLeft() != null || n.getRight() != null)
    {
      toFullString(n.getLeft(), sb);   
      toFullString(n.getRight(), sb);
    }
  }


  /**
   * Tests the BST.
   */  
    public static void main(String[] args) {
    Collection collection = new ArrayList();
    collection.add(15);
    collection.add(12);
    collection.add(18);
    collection.add(10);
    collection.add(16);
    collection.add(20);
    collection.add(11);
    collection.add(17);
    BinarySearchTree bst = new BinarySearchTree(collection);
    //System.out.println(bst);      
    String temp = bst.toFullString();
    System.out.println(temp);
    }
}

任何有关递归toFullString方法的帮助将不胜感激 .