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package algs33;
import stdlib.*;
import algs13.Queue;
/* ***********************************************************************
* Compilation: javac RedBlackLiteBST.java
* Execution: java RedBlackLiteBST < input.txt
* Dependencies: StdIn.java StdOut.java
* Data files: http://algs4.cs.princeton.edu/33balanced/tinyST.txt
*
* A symbol table implemented using a left-leaning red-black BST.
* This is the 2-3 version.
*
* This implementation implements only put, get, and contains.
* See RedBlackBST.java for a full implementation including delete.
*
*
* % more tinyST.txt
* S E A R C H E X A M P L E
*
* % java RedBlackLiteBST < tinyST.txt
* A 8
* C 4
* E 12
* H 5
* L 11
* M 9
* P 10
* R 3
* S 0
* X 7
*
*************************************************************************/
public class XRedBlackLiteBST<K extends Comparable<? super K>, V> {
private static final boolean RED = true;
private static final boolean BLACK = false;
private Node<K,V> root; // root of the BST
private int N; // number of key-value pairs in BST
// BST helper node data type
private static class Node<K,V> {
public final K key; // key
public V val; // associated data
public Node<K,V> left, right; // links to left and right subtrees
public boolean color; // color of parent link
public Node(K key, V val, boolean color) {
this.key = key;
this.val = val;
this.color = color;
}
}
/* ***********************************************************************
* Standard BST search
*************************************************************************/
// return value associated with the given key, or null if no such key exists
public V get(K key) { return get(root, key); }
public V get(Node<K,V> x, K 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 in the symbol table with the given key?
public boolean contains(K key) {
return (get(key) != null);
}
/* ***********************************************************************
* Red-black insertion
*************************************************************************/
public void put(K key, V val) {
root = insert(root, key, val);
root.color = BLACK;
assert check();
}
private Node<K,V> insert(Node<K,V> h, K key, V val) {
if (h == null) {
N++;
return new Node<>(key, val, RED);
}
int cmp = key.compareTo(h.key);
if (cmp < 0) h.left = insert(h.left, key, val);
else if (cmp > 0) h.right = insert(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);
return h;
}
/* ***********************************************************************
* red-black tree helper functions
*************************************************************************/
// is node x red (and non-null) ?
private boolean isRed(Node<K,V> x) {
if (x == null) return false;
return (x.color == RED);
}
// rotate right
private Node<K,V> rotateRight(Node<K,V> h) {
assert (h != null) && isRed(h.left);
Node<K,V> x = h.left;
h.left = x.right;
x.right = h;
x.color = h.color;
h.color = RED;
return x;
}
// rotate left
private Node<K,V> rotateLeft(Node<K,V> h) {
assert (h != null) && isRed(h.right);
Node<K,V> x = h.right;
h.right = x.left;
x.left = h;
x.color = h.color;
h.color = RED;
return x;
}
// precondition: two children are red, node is black
// postcondition: two children are black, node is red
private void flipColors(Node<K,V> h) {
assert !isRed(h) && isRed(h.left) && isRed(h.right);
h.color = RED;
h.left.color = BLACK;
h.right.color = BLACK;
}
/* ***********************************************************************
* Utility functions
*************************************************************************/
// return number of key-value pairs in symbol table
public int size() { return N; }
// is the symbol table empty?
public boolean isEmpty() { return N == 0; }
// height of tree (empty tree height = 0)
public int height() { return height(root); }
private int height(Node<K,V> x) {
if (x == null) return 0;
return 1 + Math.max(height(x.left), height(x.right));
}
// return the smallest key; null if no such key
public K min() { return min(root); }
private K min(Node<K,V> x) {
K key = null;
while (x != null) {
key = x.key;
x = x.left;
}
return key;
}
// return the largest key; null if no such key
public K max() { return max(root); }
private K max(Node<K,V> x) {
K key = null;
while (x != null) {
key = x.key;
x = x.right;
}
return key;
}
/* *********************************************************************
* Iterate using an inorder traversal.
* Iterating through N elements takes O(N) time.
***********************************************************************/
public Iterable<K> keys() {
Queue<K> queue = new Queue<>();
keys(root, queue);
return queue;
}
private void keys(Node<K,V> x, Queue<K> queue) {
if (x == null) return;
keys(x.left, queue);
queue.enqueue(x.key);
keys(x.right, queue);
}
/* ***********************************************************************
* Check integrity of red-black BST data structure
*************************************************************************/
private boolean check() {
if (!isBST()) StdOut.println("Not in symmetric order");
if (!is23()) StdOut.println("Not a 2-3 tree");
if (!isBalanced()) StdOut.println("Not balanced");
return isBST() && is23() && isBalanced();
}
// does this binary tree satisfy symmetric order?
// Note: this test also ensures that data structure 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 all keys strictly between min and max
// (if min or max is null, treat as empty constraint)
// Credit: Bob Dondero's elegant solution
private boolean isBST(Node<K,V> x, K min, K max) {
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);
}
// 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<K,V> 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 same number of black edges?
private boolean isBalanced() {
int black = 0; // number of black links on path from root to min
Node<K,V> x = root;
while (x != null) {
if (!isRed(x)) black++;
x = x.left;
}
return isBalanced(root, black);
}
// does every path from the root to a leaf have the given number of black links?
private boolean isBalanced(Node<K,V> x, int black) {
if (x == null) return black == 0;
if (!isRed(x)) black--;
return isBalanced(x.left, black) && isBalanced(x.right, black);
}
/* ***********************************************************************
* Test client
*************************************************************************/
public static void main(String[] args) {
String test = "S E A R C H E X A M P L E";
String[] keys = test.split(" ");
XRedBlackLiteBST<String, Integer> st = new XRedBlackLiteBST<>();
for (int i = 0; i < keys.length; i++)
st.put(keys[i], i);
StdOut.println("size = " + st.size());
StdOut.println("min = " + st.min());
StdOut.println("max = " + st.max());
StdOut.println();
// print keys in order using allKeys()
StdOut.println("Testing keys()");
StdOut.println("--------------------------------");
for (String s : st.keys())
StdOut.println(s + " " + st.get(s));
StdOut.println();
// insert N elements in order if one command-line argument supplied
if (args.length == 0) return;
int N = Integer.parseInt(args[0]);
XRedBlackLiteBST<Integer, Integer> st2 = new XRedBlackLiteBST<>();
for (int i = 0; i < N; i++) {
st2.put(i, i);
int h = st2.height();
StdOut.println("i = " + i + ", height = " + h + ", size = " + st2.size());
}
StdOut.println("size = " + st2.size());
}
}
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