A binary search tree (BST) is a binary tree where each node has a Comparable key (and an associated value) and satisfies the restriction that the key in any node is larger than the keys in all nodes in that node’s left subtree and smaller than the keys in all nodes in that node’s right subtree.
Representation. We define a private nested class to define nodes in BSTs, just as we did for linked lists. Each node contains a key, a value, a left link, a right link, and a node count (when relevant, we include node counts in red above the node in our drawings). The left link points to a BST for items with smaller keys, and the right link points to a BST for items with larger keys. The instance variable N gives the node count in the subtree rooted at the node. This field facilitates the implementation of various ordered symbol-table operations, as you will see. The private size() method in below algorithm is implemented to assign the value 0 to null links, so that we can maintain this field by making sure that the invariant:
size(x) = size(x.left) + size(x.right) + 1
holds for every node x in the tree.
A BST represents a set of keys (and associated values), and there are many different BSTs that represent the same set. If we project the keys in a BST such that all keys in each node’s left subtree appear to the left of the key in the node and all keys in each node’s right subtree appear to the right of the key in the node, then we always get the keys in sorted order. We take advantage of the flexibility inherent in having many BSTs represent this sorted order to develop efficient algorithms for building and using BSTs.
public class BST<Key extends Comparable<Key>, Value> {
private Node root; // root of BST
private class Node {
private Key key; // sorted by key
private Value val; // associated data
private Node left, right; // left and right subtrees
private int N; // number of nodes in subtree
public Node(Key key, Value val, int N) {
this.key = key;
this.val = val;
this.N = N;
}
}
// is the symbol table empty?
public boolean isEmpty() {
return size() == 0;
}
// return number of key-value pairs in BST
public int size() {
return size(root);
}
// return number of key-value pairs in BST rooted at x
private int size(Node x) {
if (x == null) return 0;
else return x.N;
}
/**
* ********************************************************************
* Search BST for given key, and return associated value if found,
* return null if not found
* *********************************************************************
*/
// does there exist a key-value pair with given key?
public boolean contains(Key key) {
return get(key) != null;
}
// return value associated with the given key, or null if no such key exists
public Value get(Key key) {
return get(root, key);
}
private Value get(Node x, Key key) {
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp < 0) return get(x.left, key);
else if (cmp > 0) return get(x.right, key);
else return x.val;
}
/**
* ********************************************************************
* Insert key-value pair into BST
* If key already exists, update with new value
* *********************************************************************
*/
public void put(Key key, Value val) {
if (val == null) {
delete(key);
return;
}
root = put(root, key, val);
assert check();
}
private Node put(Node x, Key key, Value val) {
if (x == null) return new Node(key, val, 1);
int cmp = key.compareTo(x.key);
if (cmp < 0) x.left = put(x.left, key, val);
else if (cmp > 0) x.right = put(x.right, key, val);
else x.val = val;
x.N = 1 + size(x.left) + size(x.right);
return x;
}
/**
* ********************************************************************
* Delete
* *********************************************************************
*/
public void deleteMin() {
if (isEmpty()) throw new NoSuchElementException("Symbol table underflow");
root = deleteMin(root);
assert check();
}
private Node deleteMin(Node x) {
if (x.left == null) return x.right;
x.left = deleteMin(x.left);
x.N = size(x.left) + size(x.right) + 1;
return x;
}
public void deleteMax() {
if (isEmpty()) throw new NoSuchElementException("Symbol table underflow");
root = deleteMax(root);
assert check();
}
private Node deleteMax(Node x) {
if (x.right == null) return x.left;
x.right = deleteMax(x.right);
x.N = size(x.left) + size(x.right) + 1;
return x;
}
public void delete(Key key) {
root = delete(root, key);
assert check();
}
private Node delete(Node x, Key key) {
if (x == null) return null;
int cmp = key.compareTo(x.key);
if (cmp < 0) x.left = delete(x.left, key);
else if (cmp > 0) x.right = delete(x.right, key);
else {
if (x.right == null) return x.left;
if (x.left == null) return x.right;
Node t = x;
x = min(t.right);
x.right = deleteMin(t.right);
x.left = t.left;
}
x.N = size(x.left) + size(x.right) + 1;
return x;
}
/**
* ********************************************************************
* Min, max, floor, and ceiling
* *********************************************************************
*/
public Key min() {
if (isEmpty()) return null;
return min(root).key;
}
private Node min(Node x) {
if (x.left == null) return x;
else return min(x.left);
}
public Key max() {
if (isEmpty()) return null;
return max(root).key;
}
private Node max(Node x) {
if (x.right == null) return x;
else return max(x.right);
}
public Key floor(Key key) {
Node x = floor(root, key);
if (x == null) return null;
else return x.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) {
Node x = ceiling(root, key);
if (x == null) return null;
else return x.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) {
Node t = ceiling(x.left, key);
if (t != null) return t;
else return x;
}
return ceiling(x.right, key);
}
/**
* ********************************************************************
* Rank and selection
* *********************************************************************
*/
public Key select(int k) {
if (k < 0 || k >= size()) return null;
Node x = select(root, k);
return x.key;
}
// Return key of rank k.
private Node select(Node x, int k) {
if (x == null) return null;
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) {
return rank(key, root);
}
// Number of keys in the subtree less than x.key.
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.
* *********************************************************************
*/
public Iterable<Key> keys() {
return keys(min(), max());
}
public Iterable<Key> keys(Key lo, Key hi) {
Queue<Key> queue = new Queue<Key>();
keys(root, queue, lo, hi);
return 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.compareTo(hi) > 0) return 0;
if (contains(hi)) return rank(hi) - rank(lo) + 1;
else return rank(hi) - rank(lo);
}
// height of this BST (one-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));
}
// level order traversal
public Iterable<Key> levelOrder() {
Queue<Key> keys = new Queue<Key>();
Queue<Node> queue = new Queue<Node>();
queue.enqueue(root);
while (!queue.isEmpty()) {
Node x = queue.dequeue();
if (x == null) continue;
keys.enqueue(x.key);
queue.enqueue(x.left);
queue.enqueue(x.right);
}
return keys;
}
/**
* **********************************************************************
* Check integrity of BST 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");
return isBST() && isSizeConsistent() && isRankConsistent();
}
// 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;
}
}
The most difficult BST operation to implement is the delete() method that removes a key-value pair from the symbol table. We can proceed in a similar manner with deleteMin() to delete any node that has one child (or no children), but what can we do to delete a node that has two children? We are left with two links, but have a place in the parent node for only one of them. An answer to this dilemma, first proposed by T. Hibbard in 1962, is to delete a node x by replacing it with its successor. Because x has a right child, its successor is the node with the smallest key in its right subtree. The replacement preserves order in the tree because there are no keys between x.key and the successor’s key. We can accomplish the task of replacing x by its successor in four (!) easy steps:
■ Save a link to the node to be deleted in t.
■ Set x to point to its successor min(t.right).
■ Set the right link of x (which is supposed to point to the BST containing all the keys larger than x.key) to deleteMin(t.right), the link to the BST containing all the keys that are larger than x.key after the deletion.
■ Set the left link of x (which was null) to t.left (all the keys that are less than both the deleted key and its successor).
Our standard recursive setup accomplishes, after the recursive calls, the task of setting the appropriate link in the parent and decrementing the node counts in the nodes on the path to the root (again, we accomplish the task of updating the counts by setting the counts in each node on the search path to be one plus the sum of the counts in its children). While this method does the job, it has a flaw that might cause performance problems in some practical situations. The problem is that the choice of using the successor is arbitrary and not symmetric. Why not use the predecessor? In practice, it is worthwhile to choose at random between the predecessor and the successor.
BSTs are not difficult to implement and can provide fast search and insert for practical applications of all kinds if the key insertions are well-approximated by the random-key model. For our examples (and for many practical applications) BSTs make the difference between being able to accomplish the task at hand and not being able to do so. Moreover, many programmers choose BSTs for symbol-table implementations because they also support fast rank, select, delete, and range query operations. Still, as we have emphasized, the bad worst-case performance of BSTs may not be tolerable in some situations. Good performance of the basic BST implementation is dependent on the keys being sufficiently similar to random keys that the tree is not likely to contain many long paths. With quicksort, we were able to randomize; with a symbol-table API, we do not have that freedom, because the client controls the mix of operations. Indeed, the worst-case behavior is not unlikely in practice—it arises when a client inserts keys in order or in reverse order, a sequence of operations that some client certainly might attempt in the absence of any explicit warnings to avoid doing so. This possibility is a primary reason to seek better algorithms and data structures.