使用这个BST类，我能够打印出字符串，但我无法弄清楚如何按字母顺序打印它们。请帮忙。我需要在这个BST类中创建一个方法，其中元素按字母顺序打印出来。感谢。
public class BinarySearchTree<AnyType extends Comparable<? super AnyType>>
{
/**
* Construct the tree.
*/
private BinaryNode<AnyType> root;
public BinarySearchTree( )
{
root = null;
}
/**
* Insert into the tree; duplicates are ignored.
* @param x the item to insert.
*/
public void insert( AnyType x )
{
root = insert( x, root );
}
/**
* Remove from the tree. Nothing is done if x is not found.
* @param x the item to remove.
*/
public void remove( AnyType x )
{
root = remove( x, root );
}
/**
* Find the smallest item in the tree.
* @return smallest item or null if empty.
*/
public AnyType findMin( )
{
if( isEmpty( ) )
throw new UnderflowException("No elements found");
return findMin( root ).element;
}
/**
* Find the largest item in the tree.
* @return the largest item of null if empty.
*/
public AnyType findMax( )
{
if( isEmpty( ) )
throw new UnderflowException("No elements found");
return findMax( root ).element;
}
/**
* Find an item in the tree.
* @param x the item to search for.
* @return true if not found.
*/
public boolean contains( AnyType x )
{
return contains( x, root );
}
/**
* Make the tree logically empty.
*/
public void makeEmpty( )
{
root = null;
}
/**
* Test if the tree is logically empty.
* @return true if empty, false otherwise.
*/
public boolean isEmpty( )
{
return root == null;
}
/**
* Print the tree contents in sorted order.
*/
public void printTree( )
{
if( isEmpty( ) )
System.out.println( "Empty tree" );
else
printTree( root );
}
/**
* Internal method to insert into a subtree.
* @param x the item to insert.
* @param t the node that roots the subtree.
* @return the new root of the subtree.
*/
private BinaryNode<AnyType> insert( AnyType x, BinaryNode<AnyType> t )
{
if( t == null )
return new BinaryNode<AnyType>( x, null, null );
int compareResult = x.compareTo( t.element );
if( compareResult < 0 )
t.left = insert( x, t.left ); // if new item < x --> insert to its left
else if( compareResult > 0 )
t.right = insert( x, t.right ); //if new item > x ---> insert to its right
else
; // Duplicate; do nothing
return t;
}
/**
* Internal method to remove from a subtree.
* @param x the item to remove.
* @param t the node that roots the subtree.
* @return the new root of the subtree.
*/
private BinaryNode<AnyType> remove( AnyType x, BinaryNode<AnyType> t )
{
if( t == null )
return t; // Item not found; do nothing
int compareResult = x.compareTo( t.element );
if( compareResult < 0 )
t.left = remove( x, t.left );
else if( compareResult > 0 )
t.right = remove( x, t.right );
else if( t.left != null && t.right != null ) // Two children
{
t.element = findMin( t.right ).element;
t.right = remove( t.element, t.right );
}
else
t = ( t.left != null ) ? t.left : t.right;
return t;
}
/**
* Internal method to find the smallest item in a subtree.
* @param t the node that roots the subtree.
* @return node containing the smallest item.
*/
private BinaryNode<AnyType> findMin( BinaryNode<AnyType> t )
{
if( t == null )
return null;
else if( t.left == null )
return t;
return findMin( t.left );
}
/**
* Internal method to find the largest item in a subtree.
* @param t the node that roots the subtree.
* @return node containing the largest item.
*/
private BinaryNode<AnyType> findMax( BinaryNode<AnyType> t )
{
if( t != null )
while( t.right != null )
t = t.right;
return t;
}
/**
* Internal method to find an item in a subtree.
* @param x is item to search for.
* @param t the node that roots the subtree.
* @return node containing the matched item.
*/
private boolean contains( AnyType x, BinaryNode<AnyType> t )
{
if( t == null )
return false;
int compareResult = x.compareTo( t.element );
if( compareResult < 0 )
return contains( x, t.left );
else if( compareResult > 0 )
return contains( x, t.right );
else
return true; // Match
}
/**
* Internal method to print a subtree in sorted order.
* @param t the node that roots the subtree.
*/
private void printTree( BinaryNode<AnyType> t )
{
if( t != null )
{
printTree( t.left );
System.out.println( t.element );
printTree( t.right );
}
}
/**
* Internal method to compute height of a subtree.
* @param t the node that roots the subtree.
*/
private int height( BinaryNode<AnyType> t )
{
if( t == null )
return -1;
else
return 1 + Math.max( height( t.left ), height( t.right ) );
}
// Node object that stores an element + left and right nodes of the same type
//*********************************************NEW CLASS*******************************************
private static class BinaryNode<AnyType>
{
// Constructors
BinaryNode( AnyType theElement )
{
this( theElement, null, null );
}
BinaryNode( AnyType theElement, BinaryNode<AnyType> lt, BinaryNode<AnyType> rt )
{
element = theElement;
left = lt;
right = rt;
}
AnyType element; // The data in the node
BinaryNode<AnyType> left; // Left child
BinaryNode<AnyType> right; // Right child
}
}
class UnderflowException extends RuntimeException
{
/**
* Construct this exception object.
* @param message the error message.
*/
public UnderflowException( String message )
{
super( message );
}
}
答案 0 :(得分：3)
当您在BST上执行in-order traversal时，它会按排序顺序为您提供节点。
类似java的伪代码中的想法：
public void printSorted(node) {
if (node == null) return;
printSorted(node.left);
System.out.println(node.value); //or any other manipulation of the node
printSorted(node.right);
}
（PS比我建议的更好的设计是使递归调用BinaryNode<AnyType>的方法，并在递归之前检查this.left == null和this.right == null，并递归它们。从上面的伪代码中修改它很容易。
答案 1 :(得分：0)
由于字母顺序和您访问树的顺序不相关，您应该像这样分解您的解决方案：