Trees: examples (Family trees)

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1 Ch 4: Trees it s a jungle out there... I think that I will never see a linked list useful as a tree; Linked lists are used by everybody, but it takes real smarts to do a tree

2 Trees: examples (Family trees) Joe Ann Mary John Mark Sue Chris

3 Trees: examples (corporate structure) President VP Sales VP Mktg... VP Finance VP Research

4 Trees: examples (your Unix file system) $HOME comp121 comp120 news mail Assn0 Assn1 Assn2 Point PQueue

5 Definitions A tree t is a finite nonempty set of elements. One of these elements is called a root, and the remaining elements (if any) are partitioned into trees that are called subtrees of t. children of the root -- root of subtree parent of an element/ grandparent/ ancestor/ descendent/... leaves: : elements with no children degree of an element: : # children degree of a tree: : max degree of any element

6 Tree definitions: examples root subtrees A B C D E F G H J K

7 Tree definitions: examples root A B C D E F G H subtrees J K

8 Tree definitions: examples parent/ children A B C D E F G H J K

9 Tree definitions: examples ancestor/ decendents A B C D E F G H J K

10 Tree definitions: examples leaves A B C D E F G H J K

11 Tree definitions: examples degrees of elements A 4 3 B C D E 0 F G H 2 J K

12 levels Tree definitions: examples level 1 A level 2 B C D E F G H level 3 J K level 4

13 Binary trees A binary tree t is a finite (possibly empty) collection of elements. When the binary tree is not empty, it has a root element and the remaining elements (if any) are partitioned into two binary trees, called the left subtree and the right subtree of t Notes: a binary tree may be empty each element has exactly two (perhaps empty) subtrees the subtrees/ / children are ordered

14 Example binary tree: an expression tree + * / a b c d (a * b) + (c / d)

15 Example binary tree: an expression tree / + d * c a b ((a * b) + c) / d)

16 Properties of binary trees A binary tree of n elements has?? edges exactly n-1 The height/ depth of a binary tree is the number of levels in it. A binary tree of height h has at least?? and at most??? elements at least h at most 2 h - 1 The height of a binary tree with n elements is at most?? and at least??? at most n at least log 2 (n+1)

17 Binary trees (review from yesterday) A binary tree t is a finite (possibly empty) collection of elements. When the binary tree is not empty, it has a root element and the remaining elements (if any) are partitioned into two binary trees, called the left subtree and the right subtree of t Notes: a binary tree may be empty each element has exactly two (perhaps empty) subtrees the subtrees/ / children are ordered

18 Properties of binary trees (review from yesterday) A binary tree of n elements has n-1 edges The height/ depth of a binary tree is the number of levels in it. A binary tree of height h has at least h and at most 2 h - 1 elements The height of a binary tree with n elements is at most n and at least log 2 (n+1)

19 Binary trees: some more definitions A FULL binary tree (Ht h ==> #-elements = 2 h -1)

20 Binary trees: some more definitions A canonical ordering A COMPLETE binary tree (levels fill in from the left)

21 Representing binary trees As an array: Use the canonical ordering -- node with canonical ordering i goes into T[i] T[i] s left child at?? T[i] s right child at?? T[i] s parent is at?? Problems with this approach: memory inefficient (works fine for full or complete binary trees)

22 Representing binary trees As a pointer to a treenode: T LC data RC

23 Representing binary trees As a pointer to a treenode: template <class T> class node{ public: }; T data; node <T> * LC; node <T> * RC; LC data RC

24 Representing binary trees As a pointer to a treenode: template <class T> class node{ public: node(t x, node<t>* t1=null, node<t>* t2=null); T data; node <T> * LC; node <T> * RC; }; {data = x; LC = t1; RC = t2;}

25 An example tree a b e c d f g h

26 An example tree LC a RC b e c d f g h

27 An example tree LC a RC LC b RC LC e RC c d f g h

28 An example tree LC a RC LC b RC LC e RC c d 0 f f RC g 0 d 0 0 d 0 0 g 0 h 0 h 0

29 Representing binary trees template <class C> class node{ public: node(c x, node<c>* t1, node<c> * t2); C data; node<c> * LC; node<c> * RC; }; b T a e c d f

30 Recursive programming on binary trees: counting elements // in the user (client) program int countelements(node<char> * T) { if (T == NULL) return 0; return ( 1 // itself + countelements(t->lc) + countelements(t->rc) ); }

31 Recursive programming on binary trees: computing depth // in the user (client) program int depth(node<char> * T) { if (T == 0) return 0; return ( 1 + max (depth depth(t->lc), depth(t->rc) (T->RC)) ); }

32 Recursive programming on binary trees: comparing trees // in the user (client) program bool identical(node<int> * T1, node<int> * T2) { if (T1 == NULL) return (T2 == NULL); if (T2 == NULL) return false; if (T1->data!= T2->data) return false; return ( identical (T1->LC, T2->LC) && identical (T1->RC, T2->RC) ); }

33 Recursive programming on binary trees: comparing trees // in the user (client) program bool what(node<int> * T1, node<int> * T2) { if (T1 == NULL) return (T2 == NULL); if (T2 == NULL) return false; } return ( what(t1->lc, T2->LC) && what(t1->rc, T2->RC) );

34 Recursive programming on binary trees: preorder traversal // in the user (client) program void preorder(node<char> * T) { if (T == 0) return; cout << T->data; preorder(t->lc); preorder(t->rc); }

35 Recursive programming on binary trees: inorder traversal // in the user (client) program void inorder Order(node<char> * T) { if (T == 0) return; inorder Order(T->LC); cout << T->data; inorder Order(T->RC); }

36 Recursive programming on binary trees: postorder traversal // in the user (client) program void postorder Order(node<char> * T) { if (T == 0) return; postorder Order(T->LC); postorder Order(T->RC); cout << T->data; }

37 Recursive programming on binary trees:??? // in the user (client) program treenode<char * mystery(node<char> * T) { if (T == 0) return NULL; return new node<char>( T->data, mystery(t->lc), mystery(t->rc) ); } mystery returns a copy

38 Recursive programming on binary trees:??? // in the user (client) program treenode<char * puzzle(node<char> * T) { if (T == 0) return NULL; return new node<char>( T->data, puzzle(t->rc), puzzle(t->lc) ); } puzzle returns a mirror-image

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