这是我正在使用的 RedBlackTree 的实现（来自 Mark Allen Weiss，数据结构
public class RedBlackTree<AnyKey extends Comparable<? super AnyKey>,
AnyValue extends Comparable<? super AnyValue>>
implements MyTreeMap<AnyKey, AnyValue>{
private static final int BLACK = 1;
private static final int RED = 0;
// The psuedo(bogus) root, has a key value of negative infinity and a right link to the real root.
private RedBlackNode<AnyKey, AnyValue> header;
// Used in place of a null link. Will always be colored black.
private RedBlackNode<AnyKey, AnyValue> nullNode;
private RedBlackNode<AnyKey, AnyValue> current;
private RedBlackNode<AnyKey, AnyValue> parent;
private RedBlackNode<AnyKey, AnyValue> grand;
private RedBlackNode<AnyKey, AnyValue> great;
public RedBlackTree( ){
nullNode = new RedBlackNode<AnyKey, AnyValue>( null, null );
nullNode.left = nullNode.right = nullNode;
header = new RedBlackNode<AnyKey, AnyValue>( null, null );
header.left = header.right = nullNode;
}
private final int compare( AnyKey theKey, RedBlackNode<AnyKey, AnyValue> t ){
if( t == header )
return 1;
else
return theKey.compareTo( t.key );
}
private void insert( AnyKey theKey, AnyValue theValue ){
current = parent = grand = header;
nullNode.key = theKey;
nullNode.value = theValue;
while( compare( theKey, current ) != 0 ){
great = grand; grand = parent; parent = current;
current = compare( theKey, current ) < 0 ?
current.left : current.right;
// Check if two red children; fix if so
if( current.left.color == RED && current.right.color == RED )
handleReorient( theKey);
}
// Insertion fails if already present
if( current != nullNode )
throw new IllegalArgumentException( theKey.toString( ) );
current = new RedBlackNode<AnyKey, AnyValue>( theKey, theValue,
nullNode, nullNode );
// Attach to parent
if( compare( theKey, parent ) < 0 )
parent.left = current;
else
parent.right = current;
handleReorient( theKey );
}
/**
* Remove from the tree.
* @param x the item to remove.
* @throws UnsupportedOperationException if called.
*/
public void remove( AnyKey x ){
}
/**
* Internal routine that is called during an insertion
* if a node has two red children. Performs flip and rotations.
* @param item the item being inserted.
*/
private void handleReorient( AnyKey key ){
// Do the color flip
current.color = RED;
current.left.color = BLACK;
current.right.color = BLACK;
if( parent.color == RED ){ // Have to rotate
grand.color = RED;
if( ( compare( key, grand ) < 0 ) !=
( compare( key, parent ) < 0 ) )
parent = rotate( key, grand ); // Start dbl rotate
current = rotate( key, great );
current.color = BLACK;
}
header.right.color = BLACK; // Make root black
}
/**
* Internal routine that performs a single or double rotation.
* Because the result is attached to the parent, there are four cases.
* Called by handleReorient.
* @param item the item in handleReorient.
* @param parent the parent of the root of the rotated subtree.
* @return the root of the rotated subtree.
*/
private RedBlackNode<AnyKey, AnyValue> rotate( AnyKey item, RedBlackNode<AnyKey, AnyValue> parent ){
if( compare( item, parent ) < 0 )
return parent.left = compare( item, parent.left ) < 0 ?
rotateWithLeftChild( parent.left ) : // LL
rotateWithRightChild( parent.left ) ; // LR
else
return parent.right = compare( item, parent.right ) < 0 ?
rotateWithLeftChild( parent.right ) : // RL
rotateWithRightChild( parent.right ); // RR
}
/**
* Rotate binary tree node with left child.
* @param k2 the node to rotate with.
*/
private static <AnyKey, AnyValue> RedBlackNode<AnyKey, AnyValue> rotateWithLeftChild( RedBlackNode<AnyKey, AnyValue> k2 ){
RedBlackNode<AnyKey, AnyValue> k1 = k2.left;
k2.left = k1.right;
k1.right = k2;
return k1;
}
@Override
public void put(AnyKey x, AnyValue y) {
RedBlackNode<AnyKey, AnyValue> bN = find(x);
if(bN == null) {
insert(x, y);
} else {
bN.value = y;
}
}
/**
* Rotate binary tree node with right child.
* @param k1 the node to rotate with.
*/
private static <AnyKey, AnyValue> RedBlackNode<AnyKey, AnyValue> rotateWithRightChild( RedBlackNode<AnyKey, AnyValue> k1 ){
RedBlackNode<AnyKey, AnyValue> k2 = k1.right;
k1.right = k2.left;
k2.left = k1;
return k2;
}
/**
* Node in the red black tree.
* @author Letian Sun
* @param <AnyKey> can be any object type
* @param <AnyValue> can be any object type
*/
private static class RedBlackNode<AnyKey, AnyValue>{
AnyKey key;
AnyValue value;
RedBlackNode<AnyKey, AnyValue> left;
RedBlackNode<AnyKey, AnyValue> right;
int color;
RedBlackNode( AnyKey theKey, AnyValue theValue ){
this( theKey, theValue, null, null );
}
RedBlackNode( AnyKey theKey, AnyValue theValue, RedBlackNode<AnyKey, AnyValue> lt
, RedBlackNode<AnyKey, AnyValue> rt ) {
key = theKey;
value = theValue;
left = lt;
right = rt;
color = RedBlackTree.BLACK;
}
public String toString() {
return "element:" + key.toString()
+ " times:" + value.toString();
}
}
我正在尝试编写作者没有编写的删除方法。
这是我到目前为止所拥有的
@Override
public void remove( AnyKey x ){
header.right = remove( x, header.right );
}
private RedBlackNode<AnyKey, AnyValue> remove( AnyKey x,
RedBlackNode<AnyKey, AnyValue> t ){
if( t == nullNode )
return t; // Item not found; do nothing
int compareResult = x.compareTo( t.key );
if( compareResult < 0 )
t.left = remove( x, t.left );
else if( compareResult > 0 )
t.right = remove( x, t.right );
else if( t.left != nullNode && t.right != nullNode ){ // Two children
t.key = findMin( t.right ).key;
t.right = remove( t.key, t.right );
}
else
t = ( t.left != nullNode ) ? t.left : t.right;
return rotate(t.key, t );
}
private RedBlackNode<AnyKey, AnyValue> findMin(RedBlackNode<AnyKey, AnyValue> t ){
if(t != null && t.left != null) {
t = findMin(t.left);
}
return t;
}
有人发现我的方法有任何问题吗？所有的逻辑对我来说都是有意义的。
当我运行 JUnit 测试用例时，
RedBlackTree testTree = new RedBlackTree<Integer,Integer>();
for(int c=0;c<1000;c++) {
testTree.put(c, c);
}
int attempt = 60;
testTree.remove(attempt);
assertEquals(attempt - 1,testTree.get(attempt - 1).intValue());
测试失败
