Priority Queues and Heaps (Ch 5.5) Huffman Coding Trees (Ch 5.6) Binary Search Trees (Ch 5.4) Lec 5: Binary Tree. Dr. Patrick Chan

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1 ata Structure hapter Biary Trees r. Patrick ha School of om puter Sciece ad Egieerig South hia Uiversity of Techolog y Recursio recursio is a procedure which calls itself The recursive procedure call must use differet argumets from the origial oe Otherwise the procedure would always get ito a ifiite loop Outlie Recursio (h.) Biary Trees (h ) Itroductio (h.) Traversal (h.) Implemetatio (h.) Biary Search Trees (h.) Priority Queues ad Heaps (h.) Huffma odig Trees (h.) Recursio as = Example: Factorial fuctio! = (-) f ( ) = f ( it factorial(it ) { if ( == ) retur ; ) retur * factorial(-); if = * factorial() factorial()= * factorial() factorial()= * factorial() factorial()= * factorial() factorial()= factorial() factorial() factorial() factorial() factorial()

2 factorial() factorial() factorial() factorial() factorial() Lec : Biar y Tree Base ase Recursive all How may iphoes the families of studets i our school have? Recursio Lec : Biar y Tree Values of the iput variables for which we perform o recursive calls are called base cases t least oe base case Every possible chai of recursive calls must evetually reach a base case Base ase alls to the curret method Each recursive call should be defied so that it makes progress towards a base case Recursive all it factorial(it ) { if ( == ) retur ; retur * factorial(-); Recursio if x > is eve if x > is odd if x = (/)* = ( /) = () = = +( /)* = (/) = () = () = (/ )* = (/) = () = = +( /)* = (/) = () = () = It is importat that we used a variable twice here rather tha callig the method twice Recursive all Base ase = = = = Lec : Biar y Tree B F E G I J H biary tree is made up of a fiite set of odes that is either empty or cosists of a ode called the together with two biary trees which are disjoit from each other ad from the Biary Tree: Itroductio Lec : Biar y Tree it power(it x, it ) { if ( == ) retur ; if ( % == ) { y = Power(x, ( - )/); retur x * y * y; { y = Power(x, /); retur y y; p( x, ) = x p( x, ( ) / ) p( x, / ) Example has two Recursive alls Recursio

3 Biary Tree: Itroductio Root epth: Level: epth: Level: epth: Level: epth: Level: epth: Level: Paret of V Siblig of V H Subtree ed at V Height is P G V R hildre of V F B J cestors of V Iteral Node Edge Leaf Node escedats of V Biary Tree omplete biary tree If the height of the tree is d, the all levels except possibly level d- are completely full The bottom level has all odes filled i from the left side B H E T H E J I G T B Biary Tree Full biary tree Each ode beig either Leaf Iteral ode with exactly two o-empty childre E H F G E H F Biary Tree Full Biary Tree Theorem Theorem The umber of leaves i a o-empty full biary tree is oe more tha the umber of iteral odes H E T B E H F G H I Leaves Iteral Nodes Leaves Iteral Nodes 9

4 Biary Tree Full Biary Tree Theorem Theorem The umber of ull poiters i a o-empty biary tree is oe more tha the umber of odes i the tree H E T B E H F G H I Leaves Iteral Nodes Null Poiters x = Nodes + = Leaves Iteral Nodes Null Poiters x = Nodes + = 9 Biary Tree: Node // Biary tre e ode class class BiNodePtr : public BiNode<Elem> { private: Elem it; // The ode's value BiNodePtr* lc; // Poiter to left child BiNodePtr* rc; // Poiter to right child public: BiNodePtr() { lc = rc = ; BiNodePtr(Elem e, BiNodePtr* l=, BiNodePtr* r=) { it = e; lc = l; rc = r; Elem& val() { retur it; void setval(cost Elem& e) { it = e; ilie BiNode<Elem>* left() co st { retur lc; ; void setleft(binode<elem>* b) { lc = (BiNodePtr*)b; ilie BiNode<Elem>* right() cost { retur rc; void setright(binode<elem>* b) { rc = (BiNodePtr*)b; bool isleaf() { retur (lc == ) && (rc == ); BiNodePtr anode(); Biary Tree: Node T Template <class Elem> class BiNode { public: virtual Elem& val( ) =; virtual BiNode* left( ) cost = ; virtual BiNode* right( ) cost = ; virtual void setval( cost Elem& ) = ; virtual void setleft( BiNode* ) = ; virtual void setright( BiNode* ) = ; virtual bool isleaf( ) = ; F G I J Biary Tree: Traversals Traversal is a process for visitig the odes i some order y traversal that lists every ode i the tree exactly oce is called a eumeratio of the tree s odes F G I J

5 Biary Tree: Traversals Preorder traversal each ode before visitig its childre B E F G H I J Postorder traversal each ode after visitig its childre B F G E J I H Iorder traversal the left subtree, the the ode, the the right subtree B F E G I J H F G I J preorder() E H I E H subtree= I.L.R B void preorder(binode<elem>* sub) { if (sub == ) retur; visit(sub); preorder(sub->left()); preorder(sub->right()); B.L.L.R Null B.R Null E H E.L H.L Null E.R Null Null H.R I I.L Null I.R Null 9 Biary Tree: Traversals Preorder traversal each ode before visitig its childre void preorder(binode<elem>* sub) { if (sub == ) retur; // Em pty visit(sub); // Perform some actios preorder(sub->left()); // left preorder(sub->right()); // right Postorder traversal each ode after visitig its childre Iorder traversal the left subtree, the the ode, the the right subtree void postorder(binode<elem>* sub) { if (sub == ) retur; // Em pty postorder(sub->left()); // left postorder(sub->right()); // right visit(sub); // Perform some actios void iorder(binode<elem>* sub) { if (sub == ) retur; // Em pty iorder(sub->left()); // left visit(sub); // Perform some actios iorder(sub->right()); // right Biary Tree: Traversals Which oe is better? () void preorder(binode<elem>* sub) { if (sub == ) retur; // Empty visit(sub); // Perform some actio preorder(sub->lc()); preorder(sub->rc()); check if the tree is empty at first F G I J () void preorder(binode<elem>* sub) { visit(sub); // Perform some actio if (sub->left()!= ) preorder(sub->lc()); if (sub->right()!= ) preorder(sub->rc()); check if the right ad left poit are empty

6 Biary Tree: Traversals Which oe is better? dvatage of ase Reduce half recursive calls (refer to Theorem i slide ) dvatage of ase Oly oe base case checkig For complex traversals, checkig criteria may be very complicated void the bugs E.g. the iitial call passes a empty tree ase should be used F G I J Biary Tree Space Requiremets Example Full tree with all odes are the same (two poiters to childre ad oe elemet): Total space required: (p + d) : umber of odes p: the amout of space required by a poiter d: the amout of space required by a data value Overhead Space: p Overhead Fractio: p / (p + d) If p = d, it meas / of total space is take up i overhead E H F G Biary Tree Space Requiremets Overhead Space mout of space ecessary to maitai the data structure i.e ay space ot used to store the data records Overhead Fractio mout of overhead space divided by amout of total space used Small Exercise!!!! Q. Write a recursive fuctio that returs the umber of odes i a biary tree it fidnumber(binode<elem>* sub) if (sub == ) retur ; // Empty subtree it L = fidnumber(sub->left()); it R = fidnumber(sub->right()); retur L+R+; F G

7 Small Exercise!!!! Q. Write a recursive fuctio that returs the umber of iteral odes i a biary tree it fiditnumber(binode<elem>* sub) if (sub == ) retur ; // Empty subtree it L = fiditnumber(sub->left()); it R = fiditnumber(sub->right()); if (sub->left()!= sub->right()!= retur L+R+; retur L+R; F G Small Exercise!!!! Q. Write a recursive fuctio that search value K i a biary tree. The fuctio returs true if value K is exist template <class Key, class Elem, class KEomp> bool search(binode<elem>* sub, Key K) if (sub == ) retur false; bool L = search(sub->left()); bool R = search(sub->right()); bool M = (sub->value() == K); retur L R M; F G Small Exercise!!!! Q. Write a recursive fuctio that returs the height of a biary tree it fidheight(binode<elem>* sub) if (sub == ) retur ; // Empty subtree it L = fidheight(sub->left()); it R = fidheight(sub->right()); if (L > R) retur L + ; retur R + ; F G Biary Search Trees Biary Search Tree (BST) is a special case of Biary Tree ll elemets stored i the left subtree of a ode with value K have values < K ll elemets stored i the right subtree of a ode with value K have values >= K

8 Biary Search Trees: T bool isert(cost Elem& e) { = iserthelp(, e); odecout++; bool remove(cost Key& K, Elem& e) { BiNode<Elem>* t = ; = removehelp(, K, t); if (t == ) retur false; e = t->val(); odecout; delete t; bool isert(cost Elem& e) { = iserthelp(, e); odecout++; Biary Search Trees: T iserthelp fuctio template <class Key, class Elem, class KEomp, class EEomp> BiNode<Elem>* BST<Key,Elem,KEomp,EEomp>:: iserthelp( BiNode<Elem>* sub, cost Elem& val ) { Base ase if (sub == ) // Empty: create ode retur ew BiNodePtr<Elem>(val,,); Recursive all if (EEomp::lt(val, sub->val())) sub->setleft(iserthelp(sub->left(), val)); Recursive all sub->setright(iserthelp(sub->right(), val)); retur sub; // Retur subtree with ode iserted =iserthelp(,); subtree= Retur Biary Search Trees: T // BST implemetatio for the ictioary T template <class Key, class Elem, class KEomp, class EEomp> class BST : public ictioary<key, Elem,KEomp, EEomp> { private: BiNode<Elem>* ; // Root of the BST it odecout; // Number of odes void clearhelp(binode<elem>*); BiNode<Elem>* iserthelp(binode<elem>*, cost Elem&); BiNode<Elem>* deletemi(binode<elem>*,binode<elem>*&); BiNode<Elem>* removehelp (BiNode<Elem>*,cost Key&,BiNode<Elem>*&); bool fidhelp(binode<elem>*, cost Key&,Elem&) cost; void prithelp(binode<elem>*, it) cost; public: BST() { = ; odecout = ; ~BST() { clearhelp(); void clear() { clearhelp(); = ; odecout = ; 9 Biary Search Trees: T bool removey(elem& e) { // elete mi value if ( == ) retur false; // Empty BiNode<Elem>* t; = deletemi(, t); e = t->val(); delete t; odecout; bool fid(cost Key& K, Elem& e) cost { retur fidhelp(, K, e); it size() { retur odecout; void prit() cost { if ( == ) cout << "The BST is empty.\"; prithelp(, );

9 Biary Search Trees: T iserthelp fuctio template <class Key, class Elem, class KEomp, class EEomp> BiNode<Elem>* BST<Key,Elem,KEomp,EEomp>:: iserthelp( BiNode<Elem>* sub, cost Elem& val ) { if (sub == ) // Empty: create ode retur ew BiNodePtr<Elem>(val,,); if (EEomp::lt(val, sub->val())) sub->setleft(iserthelp(sub->left(), val)); sub->setright(iserthelp(sub->right(), val)); retur sub; // Retur subtree with ode iserted subtree= subtree= Retur.L Retur =iserthelp(,); =iserthelp(,); bool isert(cost Elem& e) { = iserthelp(, e); odecout++; Base ase Recursive all Recursive all Biary Search Trees: T deletemi fuctio template <class Key, class Elem, class KEomp, class EEomp> BiNode<Elem>* BST<Key, Elem, KEomp, EEomp>:: deletemi(binode<elem>* sub, BiNode<Elem>*& mi) { if (sub->left() == ) { mi = sub; retur sub->right(); { // otiue left sub->setleft( deletemi(sub->left(), mi)); retur sub; Recursive all subtree= subtree=.l.r (9) Base ase ami 9 =deletemi(,ami); Biary Search Trees: T iserthelp fuctio template <class Key, class Elem, class KEomp, class EEomp> BiNode<Elem>* BST<Key,Elem,KEomp,EEomp>:: iserthelp( BiNode<Elem>* sub, cost Elem& val ) { if (sub == ) // Empty: create ode retur ew BiNodePtr<Elem>(val,,); if (EEomp::lt(val, sub->val())) sub->setleft(iserthelp(sub->left(), val)); sub->setright(iserthelp(sub->right(), val)); retur sub; // Retur subtree with ode iserted subtree= Retur.L subtree= Retur =iserthelp(,); =iserthelp(,); =iserthelp(,9);.r 9 subtree= Retur bool isert(cost Elem& e) { = iserthelp(, e); odecout++; Base ase 9 Recursive all Recursive all Biary Search Trees: T removehelp fuctio template <class Key, class Elem, class KEomp, class EEomp> BiNode<Elem>* BST<Key,Elem,KEomp,EEomp>:: removehelp(binode<elem>* sub, cost Key& K, BiNode<Elem>*& t) { if (sub == ) retur ; aot fid the target if (KEomp::lt(K, sub->val())) Searchig sub->setleft(removehelp(sub->left(), K, t)); if (KEomp::gt(K, sub->val())) sub->setright(removehelp(sub->right(), K, t)); { // Foud it: remove it Foud the target BiNode<Elem>* temp; t = sub; if (sub->left() == ) sub = sub->right(); if (sub->right() == ) sub = sub->left(); { // Both childre are o-empty sub->setright(deletemi(sub->right(), temp)); Elem te = sub->val(); sub->setval(temp->val()); temp->setval(te); t = temp; retur sub; =removehelp(,,abnode); Left Right Bigo! 99.L bool remove(cost Key& K, Elem& e) { BiNode<Elem>* t = ; = removehelp(, K, t); if (t == ) retur false; e = t->val(); odecout; delete t; te = Left Right Bigo!.L t temp sub Left Right Bigo! 99

10 Biary Search Trees: T removehelp fuctio No Left hild 9 No Right hild 9 9 Have both hildre Biary Search Trees: T prithelp fuctios template <class Key, class Elem, class KEomp, class EEomp> void BST<Key,Elem,KEomp,EEomp>:: prithelp(binode<elem>* sub, it level) { if (sub == ) retur ; prithelp(sub->left(), level+); for (it i=; i<level; i++) cout << ; cout << sub->val() << \ ; prithelp(sub->right(), level+); Base ase Recursive all Prit the ode Recursive all prithelp(, ); Biary Search Trees: T clearhelp fuctios void clear() { clearhelp(); = ; odecout = ; template <class Key, class Elem, class KEomp, class EEomp> void BST<Key,Elem,KEomp,EEomp>:: clearhelp(binode<elem>* sub) { if (sub == ) retur; Base ase clearhelp(sub->left()); Recursive all clearhelp(sub->right()); Recursive all delete sub; clearhelp(); elete the ode 9 Small Quiz!!!! Q. raw the BST that results from each of the followig commads: abst abst.isert(); abst.removey(temp); abst.removey(temp); abst.remove(, temp); abst.remove(, temp); abst.remove(9, temp); abst.remove(, temp); abst.prit();

11 Small Quiz!!!! abst.isert(); abst.removey(temp); abst.removey(temp); abst.remove(, temp); abst.remove(, temp); abst.remove(9, temp); abst.remove(, temp); abst.prit(); Biary Tree rray-based omplete BT Simple ad compact array implemetatio approach for complete biary tree No overhead space omplete biary Tree is useful: Heap data structure Exteral sortig algorithm B H E T J I G Biary Search Trees: T ost of BST Operatios ssume d = depth of the tree Fid: Θ(d) Isert: Θ(d) elete: Θ(d) d is Θ(log ) if tree is balaced The worst case: Θ() The average case: Θ(log ) Biary Tree rray-based omplete BT Paret(r) = (r - ) / if r ad r < Leftchild(r) = r + if r+ < Rightchild(r) = r + if r + < Leftsiblig(r) = r - if r is eve, r > ad r < Rightsiblig(r) = r + if r is odd, r + < 9 Positio Paret Left hild 9 Right hild Left Siblig Right Siblig 9 9 9

12 Heaps Heap is a data structure with the followig properties: omplete Biary Tree Value stored i a heap are partially ordered There is a relatioship betwee the value stored at ay ode ad values of its childre Two types of heaps Mi-heap: ll values less tha or equal to child values Max-heap: ll values greater tha or equal to child values Heap ca be implemeted simply by usig the array-based complete biary tree Heaps: T template<class Elem,class omp> class maxheap{ private: Elem* Heap; // Poiter to the heap array it size; // Maximum size of the heap it ; // Number of elems ow i heap void siftdow(it); // Put elemet i place public: maxheap(elem* h, it um, it max); it heapsize() cost; bool isleaf(it pos) cost; it leftchild(it pos) cost; it rightchild(it pos) cost; it paret(it pos) cost; bool isert(cost Elem&); bool removemax(elem&); bool remove(it, Elem&); void buildheap(); ; Heaps Max-heap Mi-heap 99 Heaps buildheap fuctio public void buildheap() // Heapify cotets { for (it i=/-; i>=; i) siftdow(i); o t eed to call siftdow o leaf odes template <class Elem, class omp> void maxheap<elem,omp>::siftdow(it pos) { while (!isleaf(pos)) { it j = leftchild(pos); it rc = rightchild(pos); if ((rc<) && omp::lt(heap[j],heap[rc])) j = rc; if (!omp::lt(heap[pos], Heap[j])) retur; swap(heap, pos, j); pos = j; Positio Paret Left hild Right hild Left Siblig Right Siblig Positio value aheap.buildheap(); aheap.siftdow(); aheap.siftdow(); aheap.siftdow(); pos j rc

13 Heaps buildheap fuctio aheap.buildheap(); aheap.siftdow(); aheap.siftdow(); aheap.siftdow(); aheap.siftdow(); aheap.siftdow(); aheap.siftdow(); 9 Heaps remove fuctio Refer to Figure., pp template <class Elem, class omp> bool maxheap<elem, omp>::remove(it pos, Elem& it) { bool flag = false; if ((pos < ) (pos >= )) retur false; swap(heap, pos, ); while ((pos!= ) && (omp::gt(heap[pos],heap[paret(pos)]))) { swap(heap, pos, paret(pos)); pos = paret(pos); flag=true; if (!flag) siftdow(pos); it = Heap[]; avalue = aheap.remove(, avalue); 9 Heaps removemax fuctio template <class Elem, class omp> bool maxheap<elem, omp>:: removemax(elem& it) { if ( == ) retur false; // Heap is empty swap(heap,, ); // Swap max with ed if (!= ) siftdow(); it = Heap[]; // Retur max value aheap.removemax(avalue); avalue = Heaps remove fuctio Refer to Figure., pp template <class Elem, class omp> bool maxheap<elem, omp>::remove(it pos, Elem& it) { bool flag = false; if ((pos < ) (pos >= )) retur false; swap(heap, pos, ); while ((pos!= ) && (omp::gt(heap[pos],heap[paret(pos)]))) { swap(heap, pos, paret(pos)); pos = paret(pos); flag=true; if (!flag) siftdow(pos); it = Heap[]; avalue = 9 aheap.remove(, avalue); 9 9

14 Small Exercise!!!! Heaps isert fuctio public void buildheap() { for (it i=/-; i>=; i) siftdow(i); ah.siftdow(); template <class Elem, class omp> // Isert elemet bool maxheap<elem, omp>::isert(cost Elem& val) { if ( >= size) retur false; // Heap is full it curr = ++; Heap[curr] = val; // Start at ed of heap // Now sift up util curr's paret > curr while ((curr!=) && (omp::gt(heap[curr], Heap[paret(curr)]))) { swap(heap, curr, paret( curr)); curr = paret( curr); 9 9 aheap.isert(9); ah.buildheap(); 9 9 ah.removemax(temp); ah.remove(, temp); ah.siftdow(); Small Exercise!!!! ah.isert(); 9 ah.siftdow(); ah.siftdow(); ah.siftdow(); 9 ah (maxheap) ah.siftdow(); ah.siftdow(); ah.siftdow(); Small Exercise!!!! ah.removemax(temp);

15 Small Exercise!!!! ah.remove(, temp); Heaps Whe a heap with size is created, there are two choices: Buildheap fuctio Isert values to a heap usig isert fuctio Which oe is better? 9 Small Exercise!!!! ah.isert(); Heaps Time omplexity Isert values to a heap Isert oe value takes Θ(log ) i the worst case Move from the bottom to the top of the tree Buildheap fuctio ( values i the array) epeds o umber of shiftdow calls = log = ( log i =Θ ( log i ) = Θ i ( ) ) x = x = x = 9 We should use buildheap to build the heap i- / i = i i

16 Heaps: Priority Queues priority queue stores objects, ad o request releases the object with greatest value Example: Schedulig jobs i a multi-taskig operatig system Huffma odig Trees etropy ecodig algorithm used for lossless data compressio Use a variable-legth code table for ecodig a source symbol erived based o the estimated probability of occurrece for each possible value Heaps: Priority Queues Possible Solutios: Liked List Isert appeds to a liked list ( O() ) Remove max value determies the maximum by scaig the list ( O() ) Sorted Liked List (decreasig order) Isert places a elemet i its correct positio ( O() ) Remove max value simply removes the head of the list ( O() ) Heap Both isert ad remove max value are O( log ) operatios Huffma odig Trees Fixed-Legth odig SII codes: bits per character E : Z : Variable-Legth odig a take advatage of relative frequecy of letters to save space Z K F U L E Use larger umber of bits to store it Use smaller umber of bits to store it

17 Huffma odig Trees Build the tree with miimum exteral path weight ombie the two smallest values each time 9 9 Z K M L U E Huffma odig Trees Huffma odig has the prefix property No code i the set is the prefix of aother No-prefix example: : B: : What is? or B More tha oe solutios ecode: Z U M E Letter Freq ode Bit E M K L U Huffma odig Trees 9 K Z M L 9 U E Letter E M K L U Z Freq ode Bit Huffma odig Trees Expected ost Expected ost per letter (how may bits have bee used for a letter i average?) x + x + x = / =. + x + x + x + x + x Letter E M K L U Z Freq ode Bit