二叉查找树在最坏的情况下性能很差时间复杂度是O(N),本节我们学习一种新的数据结构红黑树 它能保证操作在最坏情况下时间复杂度也是O(logN),无论怎么构造,它都能到达平衡，红黑树结构很复杂，首先我们学习一种叫2-3树的数据结构来帮忙对红黑树的理解。 一般的二叉查找树一个节点只能有一个键两个子节点，在2-3树中一个节点最多可以有两个键3个子节点，左子树中的键都比左边的键小， 中间子树的键大于左边的键小于右边的键，右子树的键都大于右边的键，2-3树的示意图如下
//我们结合下图来理解2-3树的构造过程，
插入的过程总是先把一个节点插入满3个键再进行分离，看右图的过程，先插入A,C,然后插入E,此时节点含3个键进行分裂，A,E各占一个节点，C移到父节点，然后插入H,L后再进行分裂，H移到父节点，然后L,M,P进行分裂，M移到父节点，这时父节点包含CHM3个键，所以父节点继续往上分裂，分裂的过程不会影响数的有序性和平衡性
从图中可以看出不论怎么构造2-3树总能达到平衡 直接用2-3树这种数据结构来达到平衡行也可以实现，但是不太方便，因为要考虑的情况太多 红黑树对2-3树的结构做了些改变，使用简单的结构就可以表达和实现2-3树 红黑树把2-3树中含有两个键的节点想象成2个用红线连接在一起的节点，2-3树中的普通链接用黑线表示，这样红黑树还是一棵二叉树，红黑树和2-3树的对应关系如下
红黑树有以下特征:
1.红链接均为左链接
2.没有节点同时和2个红链接相连，相当于含3个键的2-3树
3.红黑树是完美黑色平衡的，即每个叶子节点到根节点黑链接数量相同 //红黑树的实现过程如下:
public class RedBlackBST<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 private 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,每个节点增加一个颜色值,代表父节点指向自己链接的颜色 private int size; // subtree count
public Node(Key key, Value val, boolean color, int size) { this.key = key; this.val = val; this.color = color; this.size = size; } } /** * Initializes an empty symbol table. */ public RedBlackBST() { }
/*************************************************************************** * 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; 0 if x is null private int size(Node x) { if (x == null) return 0; return x.size; }
/** * Returns the number of key-value pairs in this symbol table. * @return the number of key-value pairs in this symbol table */ public int size() { return size(root); }
/** * Is this symbol table empty? * @return {@code true} if this symbol table is empty and {@code false} otherwise */ public boolean isEmpty() { return root == null; } /** * Does this symbol table contain the given key? * @param key the key * @return {@code true} if this symbol table contains {@code key} and * {@code false} otherwise * @throws IllegalArgumentException if {@code key} is {@code null} */ public boolean contains(Key key) { return get(key) != null; }
/*************************************************************************** * Standard BST search. ***************************************************************************/
/** * Returns the value associated with the given key. * @param key the key * @return the value associated with the given key if the key is in the symbol table * and {@code null} if the key is not in the symbol table * @throws IllegalArgumentException if {@code key} is {@code null} */ public Value get(Key key) { if (key == null) throw new IllegalArgumentException("argument to get() is null"); return get(root, key); }
// value associated with the given key in subtree 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; } /*************************************************************************** * Red-black tree insertion. ***************************************************************************/
/** * Inserts the specified key-value pair into the symbol table, overwriting the old * value with the new value if the symbol table already contains the specified key. * Deletes the specified key (and its associated value) from this symbol table * if the specified value is {@code null}. * * @param key the key * @param val the value * @throws IllegalArgumentException if {@code key} is {@code null} */ public void put(Key key, Value val) { if (key == null) throw new IllegalArgumentException("first argument to put() is null"); if (val == null) { delete(key); return; }
root = put(root, key, val); root.color = BLACK; // assert check(); } // insert the key-value pair in the subtree rooted at h //2-3树在插入过程中总是先往一个节点中插，当节点中键的个数为3时再分裂成2个含有1个键的节点，另一个键移到父节点中， //移动后如果父节点变成含3个键的节点，则再分裂父节点 //红黑树也是一样,插入的节点总是先标记为红色，然后如果不满足红黑树的特征再进行分裂转换 private Node put(Node h, Key key, Value val) { if (h == null) return new Node(key, val, RED, 1); //插入的节点先标记为红色
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; // 插入节点后可能有下面这些情况导致不满足红黑树的特征，需要把h子树进行转换，转换后可能导致父节点构成的子树不满足特征,所以每层递归都要进行转换检查，参考上图 //并且要按照情况1,2,3的顺序来检查，因为情况1转换完可能出现情况2,情况2转换完可能出现情况3 //情况1:插入节点后h子树右链接是红色，左链接是黑色,需要进行往左旋转的操作,把红色链接转到左边 if (isRed(h.right) && !isRed(h.left)) h = rotateLeft(h); //情况2:插入节点后h子树左链接是红色,左链接的左链接也是红色，需要进行往右旋转的操作，这种情况的出现类似2-3树中子节点键个数满3个向父节点分裂 //然后导致父节点也刚好满3个了,需要继续往上分裂转换 if (isRed(h.left) && isRed(h.left.left)) h = rotateRight(h); //情况3:插入节点后h子树左链接和右链接都是红色，需要进行转换颜色的操作,类型2-3树往父亲节点分裂，这种情况也可能是情况2转变过来的 if (isRed(h.left) && isRed(h.right)) flipColors(h); h.size = size(h.left) + size(h.right) + 1; return h; } //左旋转可以通过下图理解 private Node rotateLeft(Node h) { // assert (h != null) && isRed(h.right); Node x = h.right; h.right = x.left; x.left = h; x.color = h.color; x.left.color = RED; x.size = h.size; h.size = size(h.left) + size(h.right) + 1; return x; } // make a left-leaning link lean to the right private Node rotateRight(Node h) { // assert (h != null) && isRed(h.left); Node x = h.left; h.left = x.right; x.right = h; x.color = h.color; x.right.color = RED; x.size = h.size; h.size = size(h.left) + size(h.right) + 1; return x; }
// flip the colors of a node and its two children private void flipColors(Node h) { // h must have opposite color of its two 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; } /*************************************************************************** * Red-black tree deletion. ***************************************************************************/
/** * Removes the smallest key and associated value from the symbol table. * @throws NoSuchElementException if the symbol table is empty */
我们还是通过2-3树来理解红黑树的删除操作，如下图左边的图如果删除a节点会导致树失去平衡，而右边的树删除a节点还是平衡的，因为还有个x键
所以删除最小键过程中检查每个左子节点，如果发现左子节点中只含有一个键则进行分裂的逆操作，把该节点再添加一个键，这样在删除最小键后，节点不会为空，从而保持了树的平衡性。
为左子节点添加一个键有3种情况：
情况1:如果左子节点含有2个以上的键则不做操作，
情况2:如果左子节点只有一个键，而它的兄弟节点有2个以上的键，则从兄弟节点中借一个键过来，过程如下图，代码对应moveRedLeft
情况3:如果左子节点和它的兄弟节点都只有一个键，则从父节点借一个最小节点与左子节点和兄弟节点组合在一起，过程如下图，代码对应moveRedLeft里面的flipColor操作
public void deleteMin() { if (isEmpty()) throw new NoSuchElementException("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 the minimum key rooted at h private Node deleteMin(Node h) { if (h.left == null) return null;
//说明h.left只含有一个键
if (!isRed(h.left) && !isRed(h.left.left)) h = moveRedLeft(h);
h.left = deleteMin(h.left);
//最后对树进行平衡操作 return balance(h); }
// Assuming that h is red and both h.left and h.left.left // are black, make h.left or one of its children red. private Node moveRedLeft(Node h) { // assert (h != null); // assert isRed(h) && !isRed(h.left) && !isRed(h.left.left);
flipColors(h);
//平衡树的特征只有左链接可能是红色，所以兄弟节点有多个键时只能左链接是红色 if (isRed(h.right.left)) { h.right = rotateRight(h.right); h = rotateLeft(h); flipColors(h); } return h; }
/** * Removes the largest key and associated value from the symbol table. * @throws NoSuchElementException if the symbol table is empty */ public void deleteMax() { if (isEmpty()) throw new NoSuchElementException("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 the maximum key rooted at h
删除最大值得过程跟删除最小值类似，也是给右子节点增加一个键，由于红链接都在左边，所以需要执行 if (isRed(h.left)) h = rotateRight(h);通过下图来理解，右节点l和它的兄弟节点都只有一个键，所以和左子节点规则一样从父节点和兄弟节点拿一个键与l组合起来，如果是右链接是红色可以直接通过flipcolor来实现，但红链接都在左边，所以中间加了一个roteright操作，再flipcolor就可以了。
private Node deleteMax(Node h) {
//下面的判断也可以放到moveRedRight中，这样下一句可以改成 if (h.right == null) return r.left; if (isRed(h.left)) h = rotateRight(h);
//从前面红黑树的构造图可以得出最大节点的左子节点要么为空，要么是个红链接的左子节点，但通过上面的rotateRight操作后
最大节点的左子节点一定为空
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); }
/** * Removes the specified key and its associated value from this symbol table * (if the key is in this symbol table). * * @param key the key * @throws IllegalArgumentException if {@code key} is {@code null} */ public void delete(Key key) { if (key == null) throw new IllegalArgumentException("argument to delete() is null"); if (!contains(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 given key rooted at h
删除操作结合了deleteMin/deleteMax和二叉查找树的delete操作 private Node delete(Node h, Key key) { // assert get(h, key) != null;
//说明待删除节点再左子树中，需要保证左子树中每个待检测的节点都含有2个及以上键，类似deletemin
if (key.compareTo(h.key) < 0) { if (!isRed(h.left) && !isRed(h.left.left)) h = moveRedLeft(h); h.left = delete(h.left, key); }
//说明待删除节点是h节点或在右子树中，类似deletemax 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);
//如果待删除节点是h节点，进行二叉树的删除操作，从h.right中找出最小键替代h,
这个操作其实相当于待删除的键是在h.right中，所以在进行删除前需要先执行上一句的moveRedRight保证h.right节点含有2个及以上的键 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); }
// Assuming that h is red and both h.right and h.right.left // are black, make h.right or one of its children 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); flipColors(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.size = size(h.left) + size(h.right) + 1; return h; }
/*************************************************************************** * Utility functions. ***************************************************************************/
/** * Returns the height of the BST (for debugging). * @return the height of the BST (a 1-node tree has height 0) */ 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. ***************************************************************************/
/** * Returns the smallest key in the symbol table. * @return the smallest key in the symbol table * @throws NoSuchElementException if the symbol table is empty */ public Key min() { if (isEmpty()) throw new NoSuchElementException("calls min() with empty symbol table"); 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); }
/** * Returns the largest key in the symbol table. * @return the largest key in the symbol table * @throws NoSuchElementException if the symbol table is empty */ public Key max() { if (isEmpty()) throw new NoSuchElementException("calls max() with empty symbol table"); return max(root).key; }
// the largest key in the subtree rooted at x; null if no such key private Node max(Node x) { // assert x != null; if (x.right == null) return x; else return max(x.right); }
/** * Returns the largest key in the symbol table less than or equal to {@code key}. * @param key the key * @return the largest key in the symbol table less than or equal to {@code key} * @throws NoSuchElementException if there is no such key * @throws IllegalArgumentException if {@code key} is {@code null} */ public Key floor(Key key) { if (key == null) throw new IllegalArgumentException("argument to floor() is null"); if (isEmpty()) throw new NoSuchElementException("calls floor() with empty symbol table"); Node x = floor(root, key); if (x == null) return null; else return x.key; }
// the largest key in the subtree rooted at x 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; }
/** * Returns the smallest key in the symbol table greater than or equal to {@code key}. * @param key the key * @return the smallest key in the symbol table greater than or equal to {@code key} * @throws NoSuchElementException if there is no such key * @throws IllegalArgumentException if {@code key} is {@code null} */ public Key ceiling(Key key) { if (key == null) throw new IllegalArgumentException("argument to ceiling() is null"); if (isEmpty()) throw new NoSuchElementException("calls ceiling() with empty symbol table"); Node x = ceiling(root, key); if (x == null) return null; else return x.key; }
// the smallest key in the subtree rooted at 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; }
/** * Return the key in the symbol table whose rank is {@code k}. * This is the (k+1)st smallest key in the symbol table. * * @param k the order statistic * @return the key in the symbol table of rank {@code k} * @throws IllegalArgumentException unless {@code k} is between 0 and * <em>n</em>–1 */ public Key select(int k) { if (k < 0 || k >= size()) { throw new IllegalArgumentException("argument to select() is invalid: " + k); } Node x = select(root, k); return x.key; }
// the key of rank k in the subtree rooted at 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) return select(x.right, k-t-1); else return x; }
/** * Return the number of keys in the symbol table strictly less than {@code key}. * @param key the key * @return the number of keys in the symbol table strictly less than {@code key} * @throws IllegalArgumentException if {@code key} is {@code null} */ public int rank(Key key) { if (key == null) throw new IllegalArgumentException("argument to rank() is null"); return rank(key, root); }
// number of keys less than key in the subtree 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. ***************************************************************************/
/** * Returns all keys in the symbol table as an {@code Iterable}. * To iterate over all of the keys in the symbol table named {@code st}, * use the foreach notation: {@code for (Key key : st.keys())}. * @return all keys in the symbol table as an {@code Iterable} */ public Iterable<Key> keys() { if (isEmpty()) return new Queue<Key>(); return keys(min(), max()); }
/** * Returns all keys in the symbol table in the given range, * as an {@code Iterable}. * * @param lo minimum endpoint * @param hi maximum endpoint * @return all keys in the sybol table between {@code lo} * (inclusive) and {@code hi} (inclusive) as an {@code Iterable} * @throws IllegalArgumentException if either {@code lo} or {@code hi} * is {@code null} */ public Iterable<Key> keys(Key lo, Key hi) { if (lo == null) throw new IllegalArgumentException("first argument to keys() is null"); if (hi == null) throw new IllegalArgumentException("second argument to keys() is null");
Queue<Key> queue = new Queue<Key>(); // if (isEmpty() || lo.compareTo(hi) > 0) return queue; keys(root, queue, lo, hi); return queue; }
// add the keys between lo and hi in the subtree 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.enqueue(x.key); if (cmphi > 0) keys(x.right, queue, lo, hi); }
/** * Returns the number of keys in the symbol table in the given range. * * @param lo minimum endpoint * @param hi maximum endpoint * @return the number of keys in the sybol table between {@code lo} * (inclusive) and {@code hi} (inclusive) * @throws IllegalArgumentException if either {@code lo} or {@code hi} * is {@code null} */ public int size(Key lo, Key hi) { if (lo == null) throw new IllegalArgumentException("first argument to size() is null"); if (hi == null) throw new IllegalArgumentException("second argument to size() is null");
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 tree data structure. ***************************************************************************/ private boolean check() { if (!isBST()) StdOut.println("Not in symmetric order"); if (!isSizeConsistent()) StdOut.println("Subtree counts not consistent"); if (!isRankConsistent()) StdOut.println("Ranks not consistent"); if (!is23()) StdOut.println("Not a 2-3 tree"); if (!isBalanced()) StdOut.println("Not balanced"); return isBST() && isSizeConsistent() && isRankConsistent() && 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 x, Key min, Key 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); }
// are the size fields correct? private boolean isSizeConsistent() { return isSizeConsistent(root); } private boolean isSizeConsistent(Node x) { if (x == null) return true; if (x.size != 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))) return false; 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 same number of black edges? private boolean isBalanced() { int black = 0; // number of black links on path from root to 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 leaf have the given number of black links? private boolean isBalanced(Node x, int black) { if (x == null) return black == 0; if (!isRed(x)) black--; return isBalanced(x.left, black) && isBalanced(x.right, black); }
/** * Unit tests the {@code RedBlackBST} data type. * * @param args the command-line arguments */ public static void main(String[] args) { RedBlackBST<String, Integer> st = new RedBlackBST<String, Integer>(); for (int i = 0; !StdIn.isEmpty(); i++) { String key = StdIn.readString(); st.put(key, i); } for (String s : st.keys()) StdOut.println(s + " " + st.get(s)); StdOut.println(); }
}
