十二、平衡二叉树（2-3查找树、红黑二叉树)
平衡二叉树
理想情况下。我们希望可以保持二叉查找树的平衡性，总是希望树高~lgN。
2-3查找树（保证平衡）
为了保证查找树的平衡性。我们须要一些灵活性。因此在这里我们同意树中的一个节点保存多个键。确切地说，我们将一颗标准的二叉查找树中的结点称为2-结点（一个键和两条链接）。而如今我们引入3-结点（两个键和三条链接）。2-结点和3-结点中的每条链接都相应着当中保存的键所切割产生的一个区间。
一颗完美的2-3查找树中的全部空连接到根节点的距离都是同样的
查找
注意：插入要时刻保持完美平衡性(单纯的插入不添加高度，分解才会添加高度)
向2-结点插入：
向3-结点插入：
向一个父节点为2-结点的3-结点插入新键：
向一个父节点为3-结点的3-结点插入新键：
分解根节点：
分解的原因是可能出现4结点
分解情况汇总：
分析：
即使是最坏的情况，2-3树仍有较好的性能
缺点：不方便啊
红黑二叉树
基本思想是用标准的二叉查找树和一些额外的信息（替换3-结点）来表示2-3树。红链接将两个2-结点连起来构成一个3-结点。黑链接则是2-3树中的普通链接。确切地说，我们将3-结点表示为一条由左斜的红色链接。
红黑树等价含义：
红色均为左链接； 没有不论什么一个结点同一时候和两条红色链接相连； 该树是完美黑色平衡的。即随意空链接到根节点的路径上的黑链接数量同样。
红黑二叉树既是二叉查找树。也是2-3树
旋转：
由于我们在操作的过程中，可能会造成红链接出如今右面以及两条连续的红链接，所以採用旋转
左旋转
红色右链接转化为左链接
右旋：
旋转可以保证树的有序性和完美平衡性
向2结点中插入新键：
假设新键大于右键，要放在右面，所以要rotateLeft(root):
注意：插入的都是红链接
向树底部的2-结点插入新键
在树的底部插入总是用红链接连接，假设在右就旋转
向3结点插入新键
分三种情况：
注意：根节点要时刻保持为黑色。根节点的链接由红变黑后树高度会加1
向树底部的3-结点插入新键
分三种情况：
红链接的传递
相当于2-3树中的分解
在沿着插入点到根结点的路径上向上移动时在所经过的每一个结点中顺序完毕下面操作，我们就能完毕插入操作：
1.假设右子节点是红色而左子节点是黑色。进行左旋转。
2.假设左子结点是红色，而它的左子结点也是红色。进行右旋转。
3.假设左右子结点都是红色，进行颜色改变
构造轨迹：
删除最小键：
在沿着左链接向下的过程中。保证下面情况之中的一个成立：
1. 假设当前结点的左子结点不是2-结点。完毕。
2. 假设当前结点的左子节点是2-节点而它的亲兄弟结点不是2-结点，将左子结点中的兄弟结点的一个键移动到左子结点中。
3. 假设当前结点的左子结点和它的亲兄弟都是2-结点，则将左子结点、父结点中的最小键和左子结点近期的兄弟结点合并成一个4-结点，使父结点由3-结点变成2-结点或者由4-结点变成3-结点
删除后在向上分解暂时创建的4结点
这样做避免被删除位置是个2-结点(删除时不能删除一个黑链接)。2-结点被删除会变空，造成树不是平衡的（黑链接变少了），所以将遍历到的结点变成3结点或者4结点（假设不够怎么办，找兄弟结点借一个）
性质：
代码：
除deletet以及put相关的操作外都是和BST一样的。
删除那里须要通过画2-3树来理解
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
private int N; // subtree count
public Node(Key key, Value val, boolean color, int N) {
this.key = key;
this.val = val;
this.color = color;
this.N = N;
}
}
public RedBlackBST() {
}
// 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.N;
}
public int size() {
return size(root);
}
public boolean isEmpty() {
return root == null;
}
public Value get(Key key) {
if (key == null) throw new NullPointerException("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;
}
public boolean contains(Key key) {
return get(key) != null;
}
public void put(Key key, Value val) {
if (key == null) throw new NullPointerException("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
private Node put(Node h, Key key, Value val) {
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;
}
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;
if (!isRed(h.left) && !isRed(h.left.left))
h = moveRedLeft(h);
h.left = deleteMin(h.left);
return balance(h);
}
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
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);
}
public void delete(Key key) {
if (key == null) throw new NullPointerException("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
private Node delete(Node h, Key key) {
// assert get(h, key) != null;
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);
}
// 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 = 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 the left
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 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;
}
// 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);
if (isRed(h.right.left)) {
h.right = rotateRight(h.right);
h = rotateLeft(h);
flipColors(h);
}
return 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.N = size(h.left) + size(h.right) + 1;
return h;
}
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));
}
public Key min() {
if (isEmpty()) throw new NoSuchElementException("called 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);
}
public Key max() {
if (isEmpty()) throw new NoSuchElementException("called 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);
}
public Key floor(Key key) {
if (key == null) throw new NullPointerException("argument to floor() is null");
if (isEmpty()) throw new NoSuchElementException("called 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;
}
public Key ceiling(Key key) {
if (key == null) throw new NullPointerException("argument to ceiling() is null");
if (isEmpty()) throw new NoSuchElementException("called 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;
}
public Key select(int k) {
if (k < 0 || k >= size()) throw new IllegalArgumentException();
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;
}
public int rank(Key key) {
if (key == null) throw new NullPointerException("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);
}
public Iterable<Key> keys() {
if (isEmpty()) return new Queue<Key>();
return keys(min(), max());
}
public Iterable<Key> keys(Key lo, Key hi) {
if (lo == null) throw new NullPointerException("first argument to keys() is null");
if (hi == null) throw new NullPointerException("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);
}
public int size(Key lo, Key hi) {
if (lo == null) throw new NullPointerException("first argument to size() is null");
if (hi == null) throw new NullPointerException("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);
}
@SuppressWarnings("unused")
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.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))) 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);
}
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();
}
}
