CIS 121. Introduction to Trees

Transcription

1 CIS 121 Itroductio to Trees 1

2 Tree ADT Tree defiitio q A tree is a set of odes which may be empty q If ot empty, the there is a distiguished ode r, called root ad zero or more o-empty subtrees T 1, T 2, T k, each of whose roots are coected by a directed edge from r. This recursive defiitio leads to recursive tree algorithms ad tree properties beig proved by iductio. Every ode i a tree is the root of a subtree. 2

3 A Geeric Tree 3

4 Tree Termiology q Root of a subtree is a child of r. r is the paret. q All childre of a give ode are called sibligs. q A leaf (or exteral ode) has o childre. q A iteral ode is a ode with oe or more childre q A path from ode V 1 to ode V k is a sequece of odes s.t. V i is the paret of V i+1 for 1 i k. q If there is a path from V 1 to V 2, the V 1 is a acestor of V 2 ad V 2 is a descedet of V 1. 4

5 More Tree Termiology The legth of this path is the umber of edges. The legth of the path is oe less tha the umber of odes o the path ( k 1 i this example) The depth (also called level) of ay ode i a tree is the legth of the path from root to the ode. The height of a tree is the legth of the path from the root to the deepest ode i the tree. A tree with oly oe ode (the root) has height 0. 5

6 A Uix directory tree 6

7 Tree Storage A tree ode cotais: q Data Elemet q Liks to other odes Ay tree ca be represeted with the firstchild, ext-siblig implemetatio. class TreeNode { } AyType elemet; TreeNode firstchild; TreeNode extsiblig; 7

8 Pritig a Child/Siblig Tree // depth equals the umber of tabs to idet ame private void listall( it depth ) { pritname( depth ); // Prit the ame of the object if( isdirectory( ) ) } public void listall( ) { } listall( 0 ); for each file c i this directory (i.e. for each child) c.listall( depth + 1 ); What is the output whe listall( ) is used for the Uix directory tree? 8

9 K-ary Tree If we kow the maximum umber of childre each ode will have, K, we ca use a array of childre refereces i each ode. class KTreeNode { } AyType elemet; KTreeNode childre[ K ]; 9

10 Pseudocode for Pritig a K-ary Tree // depth equals the umber of tabs to idet ame private void listall( it depth ) { pritelemet( depth ); // Prit the object if( childre!= ull ) for each child c i childre array c.listall( depth + 1 ); } public void listall( ) { listall( 0 ); } 10

11 Biary Trees A special case of K-ary tree is a tree whose odes have exactly two child refereces -- biary trees. A biary tree is a rooted tree i which o ode ca have more tha two childre AND the childre are distiguished as left ad right. 11

12 The Biary Node Class private class BiaryNode<AyType> { // Costructors BiaryNode( AyType theelemet ) { this( theelemet, ull, ull ); } BiaryNode( AyType theelemet, BiaryNode<AyType> lt, BiaryNode<AyType> rt ) { elemet = theelemet; left = lt; right = rt; } } AyType elemet; // The data i the ode BiaryNode<AyType> left; // Left child referece BiaryNode<AyType> right; // Right child referece 12

13 Full Biary Tree A full biary tree is a biary tree i which every ode is a leaf or has exactly two childre. 13

14 FBT Theorem Theorem: A FBT with iteral odes has + 1 leaves (exteral odes). Proof by strog iductio o the umber of iteral odes, : Base case: q Biary Tree of oe ode (the root) has: zero iteral odes oe exteral ode (the root) Iductive Assumptio: q Assume all FBTs with iteral odes have + 1 exteral odes. 14

15 FBT Proof (cot d) Iductive Step - prove true for a tree with + 1 iteral odes (i.e. a tree with + 1 iteral odes has ( + 1) + 1 = + 2 leaves) q q q q q Let T be a FBT of iteral odes. Therefore T has + 1 leaf odes. (Iductive Assumptio) Elarge T so it has +1 iteral odes by addig two odes to some leaf. These ew odes are therefore leaf odes. Number of leaf odes icreases by 2, but the former leaf becomes iteral. So, # iteral odes becomes + 1, # leaves becomes ( + 1) =

16 Perfect Biary Tree A Perfect Biary Tree is a Full Biary Tree i which all leaves have the same depth. 16

17 PBT Theorem Theorem: The umber of odes i a PBT is 2 h+1-1, where h is height. Proof by strog iductio o h, the height of the PBT: q Notice that the umber of odes at each level is 2 l. (Proof of this is a simple iductio - left to studet as exercise). Recall that the height of the root is 0. q q Base Case: The tree has oe ode; the h = 0 ad = 1 ad 2 (h + 1) - 1 = 2 (0 + 1) 1 = = 2 1 = 1 =. Iductive Assumptio: Assume true for all PBTs with height h H. 17

18 Proof of PBT Theorem(cot) Prove true for PBT with height H+1: q Cosider a PBT with height H + 1. It cosists of a root ad two subtrees of height <= H. Sice the theorem is true for the subtrees (by the iductive assumptio sice they have height H) the PBT with height H+1 has q (2 (H+1) - 1) odes for the left subtree + (2 (H+1) - 1) odes for the right subtree + 1 ode for the root q Thus, = 2 * (2 (H+1) 1) + 1 = 2 ((H+1)+1) = 2 ((H+1)+1)

19 Complete Biary Tree A Complete Biary Tree is a biary tree i which every level is completed filled, except possibly the bottom level which is filled from left to right. 19

20 Tree Traversals Depth-First Traversals Preorder root, left subtree, right subtree Iorder left subtree, root, right subtree Postorder left subtree, right subtree, root Breadth-First Traversal Level-order each level is prited i tur 20

21 Tree Traversals Depth-first Preorder: F, B, A, D, C, E, G, I, H (root, left, right) Iorder: A, B, C, D, E, F, G, H, I (left, root, right) ß Notice the sortig! Postorder: A, C, E, D, B, H, I, G, F (left, right, root) Breadth-first Level-order: F, B, G, A, D, I, C, E, H 21

22 Costructig Trees Is it possible to recostruct a Biary Tree from just oe of its pre-order, iorder, or postorder sequeces? 22

23 Costructig Trees (cot) Give two sequeces (say pre-order ad iorder) is the tree uique? 23

24 Fidig a elemet i a Biary Tree? Retur a referece to ode cotaiig x, retur ull if x is ot foud public BiaryNode<AyType> fid(aytype x) { retur fid(root, x); } private BiaryNode<AyType> fid( BiaryNode<AyType> ode, AyType x) { BiaryNode<AyType> t = ull; // i case we do t fid it if ( ode.elemet.equals(x) ) // foud it here?? retur ode; // ot here, look i the left subtree if(ode.left!= ull) t = fid(ode.left,x); // if ot i the left subtree, look i the right subtree if ( t == ull && ode.right!= ull) t = fid(ode.right,x); } // retur referece, ull if ot foud retur t; 24

25 Biary Trees ad Recursio A Biary Tree ca have may properties q Number of leaves q Number of iterior odes q Is it a full biary tree? q Is it a perfect biary tree? q Height of the tree Each of these properties ca be determied usig a recursive fuctio. 25

26 Recursive Biary Tree Fuctio retur-type fuctio (BiaryNode<AyType> t) { // base case usually empty tree if (t == ull) retur xxxx; // determie if the ode referred to by t has the property // traverse dow the tree by recursively askig left/right // childre if their subtree has the property } retur theresult; 26

27 Is this a full biary tree? boolea isfbt (BiaryNode<AyType> t) { // base case a empty tee is a FBT } if (t == ull) retur true; // determie if this ode is full // if just oe child, retur the tree is ot full if ((t.left == ull && t.right!= ull) (t.right == ull && t.left!= ull)) retur false; // if this ode is full, ask its subtrees if they are full // if both are FBTs, the the etire tree is a FBT // if either of the subtrees is ot FBT, the the tree is ot retur isfbt( t.right ) && isfbt( t.left ); 27

28 Other Recursive Biary Tree Fuctios Cout umber of iterior odes it coutiteriornodes( BiaryNode<AyType> t); Determie the height of a biary tree. By covetio (ad for ease of codig) the height of a empty tree is -1 it height( BiaryNode<AyType> t); May others 28

29 Other Biary Tree Operatios How do we isert a ew elemet ito a biary tree? How do we remove a elemet from a biary tree? 29