4.2 Binary Search Trees

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1 Binary trees 4. Binary earc Trees Definition. BT is a binary tree in symmetric order. root a left link a subtree binary tree is eiter: mpty. rigt cild of root Two disjoint binary trees (left and rigt). null links natomy of a binary tree BTs ordered operations deletion parent of and ymmetric order. ac node as a key, and every node s key is: arger tan all keys in its left subtree. left link of maller tan all keys in its rigt subtree. keys smaller tan key 9 value associated wit keys larger tan natomy of a binary tree lgoritms in Java, 4t dition obert edgewick and Kevin Wayne opyrigt 009 October 0, 009 0:9:6 BT representation in Java BT implementation (skeleton) Java definition. BT is a reference to a root ode. public class BT<Key extends omparable<key>, Value> private ode root; ode is comprised of four fields: Key and a Value. private class ode /* see previous slide */ reference to te left and rigt subtree. smaller keys larger keys private class ode private Key key; private Value val; private ode left, rigt; public ode(key key, Value val) tis.key = key; tis.val = val; root of BT public void put(key key, Value val) /* see next slides */ public Value get(key key) /* see next slides */ BT ode key left BT wit smaller keys val public void delete(key key) /* see next slides */ rigt public Iterable<Key> iterator() /* see next slides */ BT wit larger keys Binary tree Key and Value are generic types; Key is omparable 3 4

2 BT BT : Java implementation Get. eturn value corresponding to given key, or null if no suc key. Get. eturn value corresponding to given key, or null if no suc key. successful for black nodes could matc te key is greater tan so look to te rigt is less tan so look to te left gray nodes cannot matc te key found ( it) so return value unsuccessful for T T is greater tan so look to te rigt T is less tan so look to te left link is null so T is not in tree ( miss) public Value get(key key) ode x = root; wile (x!= null) if (cmp < 0) x = x.left; else if (cmp > 0) x = x.rigt; else if (cmp == 0) return x.val; return null; unning time. roportional to dept of node. 6 BT BT : Java implementation ut. ssociate value wit key. earc for key, ten two cases: Key in tree reset value. Key not in tree add new node. ing for ends at tis null link create new node reset links on te way up ut. ssociate value wit key. public void put(key key, Value val) root = put(root, key, val); private ode put(ode x, Key key, Value val) if (x == null) return new ode(key, val); if (cmp < 0) x.left = put(x.left, key, val); else if (cmp > 0) x.rigt = put(x.rigt, key, val); else if (cmp == 0) x.val = val; return x; concise, but tricky, recursive code; read carefully! Insertion into a BT unning time. roportional to dept of node.

3 best case BT trace: standard indexing client Tree sape 0 key value 9 black nodes are accessed in 3 0 gray nodes are untouced 6 canged value 9 BT trace for standard indexing client 6 canged value worst case typical case typical case best case 4 red nodes are new best case typical case ost of / is proportional to dept of node. any BTs correspond to same set of keys. canged value key value worst case BT possibilities casedepends on order of ion. emark. Treeworst sape BT possibilities BT ion: random order BT ion: random order visualization Observation. If keys ed in random order, tree stays relatively flat. x. Insert keys in random order. 0 BT possibilities Typical BT built from random keys ( = 6)

orrespondence between BTs and quicksort partitioning BTs: matematical analysis roposition.

K I Q T roposition. [eed, 003] If keys are ed in random order, U expected eigt of tree is ~ 4.3 ln.

orrespondence is - if no duplicate keys. 3 4 T implementations: summary implementation guarantee average case ordered ops?

4 orrespondence between BTs and quicksort partitioning BTs: matematical analysis roposition. If keys are ed in random order, te expected number of compares for a / is ~ ln. f. - correspondence wit quicksort partitioning. K I Q T roposition. [eed, 003] If keys are ed in random order, U expected eigt of tree is ~ 4.3 ln. U O But Worst-case for //eigt is. (exponentially small cance wen keys are ed in random order) emark. orrespondence is - if no duplicate keys. 3 4 T implementations: summary implementation guarantee average case ordered ops? operations on keys no equals() lg / yes compareto().39 lg.39 lg? compareto() it sequential (unordered list) / binary (ordered array) lg BT BTs ordered operations deletion osts for java Frequencyounter < tale.txt using BT 6

5 inimum and maximum Floor and ceiling inimum. mallest key in table. aximum. argest key in table. Floor. argest key to a given key. eiling. mallest key to a given key. min () max floor(g) () ceiling(q) floor(d) Q. ow to find te min / max. Q. ow to find te floor /ceiling. omputing te floor omputing te floor ase. [k equals te key at root] Te floor of k is k. ase. [k is less tan te key at root] Te floor of k is in te left subtree. ase 3. [k is greater tan te key at root] Te floor of k is in te rigt subtree (if tere is any key k in rigt subtree); oterwise it is te key in te root. finding floor(g) G is greater tan so floor(g) could be on te rigt floor(g)in left subtree is null G is less tan so floor(g) must be on te left public Key floor(key key) ode x = floor(root, key); if (x == null) return null; return x.key; private ode floor(ode x, Key key) if (x == null) return null; if (cmp == 0) return x; if (cmp < 0) return floor(x.left, key); ode t = floor(x.rigt, key); if (t!= null) return t; else return x; finding floor(g) G is greater tan so floor(g) could be on te rigt floor(g)in left subtree is null G is less tan so floor(g) must be on te left result result omputing te floor function 9 omputing te floor function 0

ubtree counts BT implementation: subtree counts In eac node, we store te number of nodes in te subtree rooted at tat node. To implement size(), return te count at te root. node count 6 3 emark.

6 ubtree counts BT implementation: subtree counts In eac node, we store te number of nodes in te subtree rooted at tat node. To implement size(), return te count at te root. node count 6 3 emark. Tis facilitates efficient implementation of rank() and select(). private class ode private Key key; private Value val; private ode left; private ode rigt; private int ; nodes in subtree public int size() return size(root); private int size(ode x) if (x == null) return 0; return x.; private ode put(ode x, Key key, Value val) if (x == null) return new ode(key, val); if (cmp < 0) x.left = put(x.left, key, val); else if (cmp > 0) x.rigt = put(x.rigt, key, val); else if (cmp == 0) x.val = val; x. = + size(x.left) + size(x.rigt); return x; ank Inorder traversal ank. ow many keys < k? asy recursive algoritm (4 cases!) public int rank(key key) return rank(key, root); node count 6 3 Traverse left subtree. nqueue key. Traverse rigt subtree. public Iterable<Key> keys() Queue<Key> q = new Queue<Key>(); inorder(root, queue); return q; key BT val private int rank(key key, ode x) if (x == null) return 0; if (cmp < 0) return rank(key, x.left); else if (cmp > 0) return + size(x.left) + rank(key, x.rigt); else return size(x.left); private void inorder(ode x, Queue<Key> q) if (x == null) return; inorder(x.left, q); q.enqueue(x.key); inorder(x.rigt, q); left rigt BT wit smaller keys BT wit larger keys smaller keys, in order key larger keys, in order all keys, in order roperty. Inorder traversal of a BT yields keys in ascending order. 3 4

select ordered iteration log lg lg worst-case running time of ordered symbol table operations = eigt of BT (proportional to log if keys ed in random order) recursive calls queue function call stack 6

7 Inorder traversal BT: ordered symbol table operations summary Traverse left subtree. nqueue key. Traverse rigt subtree. sequential binary BT lg inorder() inorder() inorder() enqueue inorder() enqueue enqueue inorder() inorder() enqueue inorder() enqueue print enqueue inorder() enqueue min / max floor / ceiling rank select ordered iteration log lg lg worst-case running time of ordered symbol table operations = eigt of BT (proportional to log if keys ed in random order) recursive calls queue function call stack 6 T implementations: summary implementation guarantee delete it average case delete ordered iteration? operations on keys sequential (linked list) / / no equals() BTs ordered operations deletion binary (ordered array) lg lg / / yes compareto() BT.39 lg.39 lg??? yes compareto() ext. Deletion in BTs.

8 BT deletion: lazy approac Deleting te minimum To remove a node wit a given key: et its value to null. eave key in tree to guide es (but don't consider it equal to key). To delete te minimum key: Go left until finding a node wit a null left link. eplace tat node by its rigt link. Update subtree counts. go left until reacing null left link I delete I tombstone public void deletein() root = deletein(root); return tat node s rigt link available for garbage collection ost. O(log ') per,, and delete (if keys in random order), were ' is te number of key-value pairs ever ed in te BT. Unsatisfactory solution. Tombstone overload. private ode deletein(ode x) if (x.left == null) return x.rigt; x.left = deletein(x.left); x. = + size(x.left) + size(x.rigt); return x; update links and counts after recursive calls Deleting te minimum in a BT 9 30 ibbard deletion ibbard deletion To delete a node wit key k: for node t containing key k. To delete a node wit key k: for node t containing key k. ase 0. [0 cildren] Delete t by setting parent link to null. ase. [ cild] Delete t by replacing parent link. deleting node to delete replace wit null link available for garbage collection update counts after recursive calls deleting update counts after recursive calls node to delete replace wit cild link available for garbage collection 3 3

ibbard deletion ibbard deletion: Java implementation To delete a node wit key k: for node t containing key k. ase. [ cildren] Find successor x of t. Delete te minimum in t's rigt subtree.

9 ibbard deletion ibbard deletion: Java implementation To delete a node wit key k: for node t containing key k. ase. [ cildren] Find successor x of t. Delete te minimum in t's rigt subtree. ut x in t's spot. deleting node to delete t go rigt, ten go left until reacing null left link x for key successor min(t.rigt) reacing null left link t.left x x as no left cild but don't garbage collect x still a BT deletein(t.rigt) Deletion in a BT update links and node counts after recursive calls 33 public void delete(key key) root = delete(root, key); private ode delete(ode x, Key key) if (x == null) return null; if (cmp < 0) x.left = delete(x.left, key); else if (cmp > 0) x.rigt = delete(x.rigt, key); else if (x.rigt == null) return x.left; ode t = x; x = min(t.rigt); x.rigt = deletein(t.rigt); x.left = t.left; x. = size(x.left) + size(x.rigt) + ; return x; for key no rigt cild replace wit successor update subtree counts 34 ibbard deletion: analysis T implementations: summary Unsatisfactory solution. ot symmetric. implementation guarantee delete it average case delete ordered iteration? operations on keys sequential (linked list) binary (ordered array) / / no equals() lg lg / / yes compareto() BT.39 lg.39 lg yes compareto() oter operations also become if deletions allowed urprising consequence. Trees not random (!) sqrt() per op. ongstanding open problem. imple and efficient delete for BTs. ext lecture. Guarantee logaritmic performance for all operations. 3 36

BINARY SEARCH TREES BBM ALGORITHMS DEPT. OF COMPUTER ENGINEERING

BINARY SEARCH TREES BBM ALGORITHMS DEPT. OF COMPUTER ENGINEERING cknowledgement: The course slides are adapted from the slides prepared by. edgewick and K. Wayne of Princeton University. BB 202 - LGOIT DPT. OF OPUT NGINING BINY T BTs Ordered operations Deletion TODY