算法4第3章红黑树及习题讲解
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();
}
}