java红黑树的控制台打印
二叉搜索树这个数据结构太重要
首先被提出来的二叉平衡搜索树的方式是AVL树，然后提出来的是Tree Heap树，最后目前在实践中最为高效的红黑树。
Java中的TreeMap也是基于红黑树实现
所以想要深入学习一下红黑树
这里首先贴上google搜到的红黑树代码
/*************************************************************************
* Compilation: javacRedBlackBST.java
* Execution: java RedBlackBST< input.txt
* Dependencies: StdIn.java StdOut.java
* Datafiles: http://algs4.cs.princeton.edu/33balanced/tinyST.txt
*
* Asymbol table implemented using a left-leaning red-black BST.
* Thisis the 2-3 version.
*
* Note:commented out assertions because DrJava now enables assertions
* by default.
*
* %more tinyST.txt
* S E AR C H E X A M P L E
*
* %java RedBlackBST < tinyST.txt
* A 8
* C 4
* E 12
* H 5
* L 11
* M 9
* P 10
* R 3
* S 0
* X 7
*
*************************************************************************/
importjava.util.LinkedList;
importjava.util.NoSuchElementException;
import java.util.Queue;
importjava.util.Random;
public classRedBlackBST<Key extends Comparable<Key>, Value> {
private static final boolean RED = true;
private static final boolean BLACK = false;
private Node root; // root of the BST
// BST helper node data type
class Node {
private Key key; // key
private Value val; // associated data
private Node left, right; // links to left and right subtrees
private boolean color; // color of parent link
private int N; // subtree count
public Node(Key key, Value val, booleancolor, int N) {
this.key = key;
this.val = val;
this.color = color;
this.N = N;
}
}
/*************************************************************************
* Node helper methods
*************************************************************************/
// is node x red; false if x is null ?
private boolean isRed(Node x) {
if (x == null) return false;
return (x.color == RED);
}
// number of node in subtree rooted at x; 0if x is null
private int size(Node x) {
if (x == null) return 0;
return x.N;
}
/*************************************************************************
* Size methods
*************************************************************************/
// return number of key-value pairs in thissymbol table
public int size() { return size(root); }
// is this symbol table empty?
public boolean isEmpty() {
return root == null;
}
/*************************************************************************
* Standard BST search
*************************************************************************/
// value associated with the given key;null if no such key
public Value get(Key key) { return get(root,key); }
// value associated with the given key insubtree rooted at x; null if no such key
private Value get(Node x, Key key) {
while (x != null) {
int cmp = key.compareTo(x.key);
if (cmp < 0) x = x.left;
else if (cmp > 0) x = x.right;
else return x.val;
}
return null;
}
// is there a key-value pair with the givenkey?
public boolean contains(Key key) {
return (get(key) != null);
}
// is there a key-value pair with the givenkey in the subtree rooted at x?
private boolean contains(Node x, Key key) {
return (get(x, key) != null);
}
/*************************************************************************
* Red-black insertion
*************************************************************************/
// insert the key-value pair; overwrite theold value with the new value
// if the key is already present
public void put(Key key, Value val) {
root = put(root, key, val);
root.color = BLACK;
// assert check();
}
// insert the key-value pair in the subtreerooted at h
private Node put(Node h, Key key, Valueval) {
if (h == null) return new Node(key,val, RED, 1);
int cmp = key.compareTo(h.key);
if (cmp < 0) h.left =put(h.left, key, val);
else if (cmp > 0) h.right = put(h.right,key, val);
else h.val = val;
// fix-up any right-leaning links
if (isRed(h.right) &&!isRed(h.left)) h = rotateLeft(h);
if (isRed(h.left) && isRed(h.left.left)) h = rotateRight(h);
if (isRed(h.left) && isRed(h.right)) flipColors(h);
h.N = size(h.left) + size(h.right) + 1;
return h;
}
/*************************************************************************
* Red-black deletion
*************************************************************************/
// delete the key-value pair with theminimum key
public void deleteMin() {
if (isEmpty()) throw newNoSuchElementException("BST underflow");
// if both children of root are black,set root to red
if (!isRed(root.left) &&!isRed(root.right))
root.color = RED;
root = deleteMin(root);
if (!isEmpty()) root.color = BLACK;
// assert check();
}
// delete the key-value pair with theminimum key rooted at h
private Node deleteMin(Node h) {
if (h.left == null)
return null;
if (!isRed(h.left) &&!isRed(h.left.left))
h = moveRedLeft(h);
h.left = deleteMin(h.left);
return balance(h);
}
// delete the key-value pair with themaximum key
public void deleteMax() {
if (isEmpty()) throw newNoSuchElementException("BST underflow");
// if both children of root are black,set root to red
if (!isRed(root.left) &&!isRed(root.right))
root.color = RED;
root = deleteMax(root);
if (!isEmpty()) root.color = BLACK;
// assert check();
}
// delete the key-value pair with themaximum key rooted at h
private Node deleteMax(Node h) {
if (isRed(h.left))
h = rotateRight(h);
if (h.right == null)
return null;
if (!isRed(h.right) && !isRed(h.right.left))
h = moveRedRight(h);
h.right = deleteMax(h.right);
return balance(h);
}
// delete the key-value pair with the givenkey
public void delete(Key key) {
if (!contains(key)) {
System.err.println("symboltable does not contain " + key);
return;
}
// if both children of root are black,set root to red
if (!isRed(root.left) &&!isRed(root.right))
root.color = RED;
root = delete(root, key);
if (!isEmpty()) root.color = BLACK;
// assert check();
}
// delete the key-value pair with the givenkey rooted at h
private Node delete(Node h, Key key) {
// assert contains(h, key);
if (key.compareTo(h.key) < 0) {
if (!isRed(h.left) &&!isRed(h.left.left))
h = moveRedLeft(h);
h.left = delete(h.left, key);
}
else {
if (isRed(h.left))
h = rotateRight(h);
if (key.compareTo(h.key) == 0&& (h.right == null))
return null;
if (!isRed(h.right) &&!isRed(h.right.left))
h = moveRedRight(h);
if (key.compareTo(h.key) == 0) {
Node x = min(h.right);
h.key = x.key;
h.val = x.val;
// h.val = get(h.right,min(h.right).key);
// h.key = min(h.right).key;
h.right = deleteMin(h.right);
}
else h.right = delete(h.right,key);
}
return balance(h);
}
/*************************************************************************
* red-black tree helper functions
*************************************************************************/
// make a left-leaning link lean to theright
private Node rotateRight(Node h) {
// assert (h != null) &&isRed(h.left);
Node x = h.left;
h.left = x.right;
x.right = h;
x.color = x.right.color;
x.right.color = RED;
x.N = h.N;
h.N = size(h.left) + size(h.right) + 1;
return x;
}
// make a right-leaning link lean to theleft
private Node rotateLeft(Node h) {
// assert (h != null) &&isRed(h.right);
Node x = h.right;
h.right = x.left;
x.left = h;
x.color = x.left.color;
x.left.color = RED;
x.N = h.N;
h.N = size(h.left) + size(h.right) + 1;
return x;
}
// flip the colors of a node and its twochildren
private void flipColors(Node h) {
// h must have opposite color of itstwo children
// assert (h != null) &&(h.left != null) && (h.right != null);
// assert (!isRed(h) && isRed(h.left) && isRed(h.right))
// || (isRed(h) &&!isRed(h.left) && !isRed(h.right));
h.color = !h.color;
h.left.color = !h.left.color;
h.right.color = !h.right.color;
}
// Assuming that h is red and both h.leftand h.left.left
// are black, make h.left or one of itschildren red.
private Node moveRedLeft(Node h) {
// assert (h != null);
// assert isRed(h) &&!isRed(h.left) && !isRed(h.left.left);
flipColors(h);
if (isRed(h.right.left)) {
h.right = rotateRight(h.right);
h = rotateLeft(h);
}
return h;
}
// Assuming that h is red and both h.rightand h.right.left
// are black, make h.right or one of itschildren red.
private Node moveRedRight(Node h) {
// assert (h != null);
// assert isRed(h) &&!isRed(h.right) && !isRed(h.right.left);
flipColors(h);
if (isRed(h.left.left)) {
h = rotateRight(h);
}
return h;
}
// restore red-black tree invariant
private Node balance(Node h) {
// assert (h != null);
if (isRed(h.right)) h = rotateLeft(h);
if (isRed(h.left) &&isRed(h.left.left)) h = rotateRight(h);
if (isRed(h.left) && isRed(h.right)) flipColors(h);
h.N = size(h.left) + size(h.right) + 1;
return h;
}
/*************************************************************************
* Utility functions
*************************************************************************/
// height of tree (1-node tree has height0)
public int height() { return height(root);}
private int height(Node x) {
if (x == null) return -1;
return 1 + Math.max(height(x.left),height(x.right));
}
/*************************************************************************
* Ordered symbol table methods.
*************************************************************************/
//the smallest key; null if no such key
public Key min() {
if (isEmpty()) return null;
return min(root).key;
}
// the smallest key in subtree rooted at x;null if no such key
private Node min(Node x) {
// assert x != null;
if (x.left == null) return x;
else return min(x.left);
}
// the largest key; null if no such key
public Key max() {
if (isEmpty()) return null;
return max(root).key;
}
// the largest key in the subtree rooted atx; null if no such key
private Node max(Node x) {
// assert x != null;
if (x.right == null) return x;
else return max(x.right);
}
// the largest key less than or equal tothe given key
public Key floor(Key key) {
Node x = floor(root, key);
if (x == null) return null;
else return x.key;
}
// the largest key in the subtree rooted atx less than or equal to the given key
private Node floor(Node x, Key key) {
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp == 0) return x;
if (cmp < 0) return floor(x.left, key);
Node t = floor(x.right, key);
if (t != null) return t;
else return x;
}
// the smallest key greater than or equalto the given key
public Key ceiling(Key key) {
Node x = ceiling(root, key);
if (x == null) return null;
else return x.key;
}
// the smallest key in the subtree rootedat x greater than or equal to the given key
private Node ceiling(Node x, Key key){
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp == 0) return x;
if (cmp > 0) return ceiling(x.right, key);
Node t = ceiling(x.left, key);
if (t != null) return t;
else return x;
}
// the key of rank k
public Key select(int k) {
if (k < 0 || k >= size()) return null;
Node x = select(root, k);
return x.key;
}
// the key of rank k in the subtree rootedat x
private Node select(Node x, int k) {
// assert x != null;
// assert k >= 0 && k <size(x);
int t = size(x.left);
if (t > k) return select(x.left, k);
else if (t < k) returnselect(x.right, k-t-1);
else return x;
}
// number of keys less than key
public int rank(Key key) {
return rank(key, root);
}
// number of keys less than key in thesubtree rooted at x
private int rank(Key key, Node x) {
if (x == null) return 0;
int cmp = key.compareTo(x.key);
if (cmp < 0) return rank(key, x.left);
else if (cmp > 0) return 1 +size(x.left) + rank(key, x.right);
else return size(x.left);
}
/***********************************************************************
* Range count and range search.
***********************************************************************/
// all of the keys, as an Iterable
public Iterable<Key> keys() {
return keys(min(), max());
}
// the keys between lo and hi, as anIterable
public Iterable<Key> keys(Key lo, Keyhi) {
Queue<Key> queue = newLinkedList<Key>();
// if (isEmpty() || lo.compareTo(hi)> 0) return queue;
keys(root, queue, lo, hi);
return queue;
}
// add the keys between lo and hi in thesubtree rooted at x
// to the queue
private void keys(Node x, Queue<Key>queue, Key lo, Key hi) {
if (x == null) return;
int cmplo = lo.compareTo(x.key);
int cmphi = hi.compareTo(x.key);
if (cmplo < 0) keys(x.left, queue,lo, hi);
if (cmplo <= 0 && cmphi>= 0) queue.add(x.key);
if (cmphi > 0) keys(x.right, queue,lo, hi);
}
// number keys between lo and hi
public int size(Key lo, Key hi) {
if (lo.compareTo(hi) > 0) return 0;
if (contains(hi)) return rank(hi) -rank(lo) + 1;
else return rank(hi) - rank(lo);
}
/*************************************************************************
* Check integrity of red-black BST data structure
*************************************************************************/
private boolean check() {
if (!isBST()) System.out.println("Not insymmetric order");
if (!isSizeConsistent())System.out.println("Subtree counts not consistent");
if (!isRankConsistent())System.out.println("Ranks not consistent");
if (!is23()) System.out.println("Not a 2-3tree");
if (!isBalanced()) System.out.println("Notbalanced");
return isBST() &&isSizeConsistent() && isRankConsistent() && is23() &&isBalanced();
}
// does this binary tree satisfy symmetricorder?
// Note: this test also ensures that datastructure is a binary tree since order is strict
private boolean isBST() {
return isBST(root, null, null);
}
// is the tree rooted at x a BST with allkeys strictly between min and max
// (if min or max is null, treat as emptyconstraint)
// Credit: Bob Dondero's elegant solution
private boolean isBST(Node x, Key min, Keymax) {
if (x == null) return true;
if (min != null &&x.key.compareTo(min) <= 0) return false;
if (max != null &&x.key.compareTo(max) >= 0) return false;
return isBST(x.left, min, x.key)&& isBST(x.right, x.key, max);
}
// are the size fields correct?
private boolean isSizeConsistent() { returnisSizeConsistent(root); }
private boolean isSizeConsistent(Node x) {
if (x == null) return true;
if (x.N != size(x.left) + size(x.right)+ 1) return false;
return isSizeConsistent(x.left)&& isSizeConsistent(x.right);
}
// check that ranks are consistent
private boolean isRankConsistent() {
for (int i = 0; i < size(); i++)
if (i != rank(select(i))) returnfalse;
for (Key key : keys())
if (key.compareTo(select(rank(key))) !=0) return false;
return true;
}
// Does the tree have no red right links,and at most one (left)
// red links in a row on any path?
private boolean is23() { return is23(root);}
private boolean is23(Node x) {
if (x == null) return true;
if (isRed(x.right)) return false;
if (x != root && isRed(x)&& isRed(x.left))
return false;
return is23(x.left) &&is23(x.right);
}
// do all paths from root to leaf have samenumber of black edges?
private boolean isBalanced() {
int black = 0; // number of black links on path from rootto min
Node x = root;
while (x != null) {
if (!isRed(x)) black++;
x = x.left;
}
return isBalanced(root, black);
}
// does every path from the root to a leafhave the given number of black links?
private boolean isBalanced(Node x, intblack) {
if (x == null) return black == 0;
if (!isRed(x)) black--;
return isBalanced(x.left, black)&& isBalanced(x.right, black);
}
/*****************************************************************************
* Print Method
*****************************************************************************/
public void padding ( String ch, int n )
{
int i;
for ( i = 0; i < n; i++ )
System.out.printf(ch);
}
void print_node (Node root, int level )
{
if ( root == null )
{
padding ( "\t", level );
System.out.println( "NIL" );
}
else
{
print_node ( root.right, level + 1 );
padding ( "\t", level );
if(root.color == BLACK)
{
System.out.printf("(%d)\n", root.key );
}
else
System.out.printf( "%d\n",root.key );
print_node ( root.left, level + 1 );
}
}
void print_tree()
{
print_node(this.root,0);
System.out.printf("-------------------------------------------\n");
}
/*****************************************************************************
* Test client
*****************************************************************************/
public static void main(String[] args) {
RedBlackBST<Integer, String> st =new RedBlackBST<Integer, String>();
/*for (int i = 0; !StdIn.isEmpty();i++) {
String key = StdIn.readString();
st.put(key, i);
}*/
Random rm=new Random();
for(int i=0;i<15;i++)
{
Strings=""+(char)('a'+Math.random()*('z'-'a'+1));
st.put(rm.nextInt(60),s);
}
for (Integer i:st.keys())
System.out.println(i + "" + st.get(i));
System.out.println();
st.print_tree();
}
}
为了便于调试,将树打印出来很重要
这里参考
http://blog.chinaunix.net/uid-24774106-id-3440620.html
其中包括了为这个java版本的红黑树添加了一下打印方法
觉得这个方法很不错,利用递归,先打印右子树,再打印当前节点,最后打印左子树,
巧妙的将树横向打印了出来
/*****************************************************************************
* Print Method
*****************************************************************************/
public void padding ( String ch, int n )
{
int i;
for ( i = 0; i < n; i++ )
System.out.printf(ch);
}
void print_node (Node root, int level )
{
if ( root == null )
{
padding ( "\t", level );
System.out.println( "NIL" );
}
else
{
print_node ( root.right, level + 1 );
padding ( "\t", level );
if(root.color == BLACK)
{
System.out.printf("(%d)\n", root.key );
}
else
System.out.printf( "%d\n",root.key );
print_node ( root.left, level + 1 );
}
}
void print_tree()
{
print_node(this.root,0);
System.out.printf("-------------------------------------------\n");
}
main方法测试
public static void main(String[] args) {
RedBlackBST<Integer, String> st =new RedBlackBST<Integer, String>();
/*for (int i = 0; !StdIn.isEmpty();i++) {
String key = StdIn.readString();
st.put(key, i);
}*/
Random rm=new Random();
for(int i=0;i<15;i++)
{
Strings=""+(char)('a'+Math.random()*('z'-'a'+1));
st.put(rm.nextInt(60),s);
}
for (Integer i:st.keys())
System.out.println(i + "" + st.get(i));
System.out.println();
st.print_tree();
}
}
运行结果如下