CS 1031 Trees A Quick Introduction to Graphs Definition of Trees Rooted Trees Binary Trees Binary Search Trees

CS 1032 Introduction to Graphs A graph is a finite set of nodes with edges between nodes Formally, a graph G is a structure (V,E) consisting of –a finite set V called the set of nodes, and –a set E that is a subset of VxV. That is, E is a set of pairs of the form (x,y) where x and y are nodes in V

CS 1033 Examples of Graphs V={1,2,3,4,5} E={(1,2), (2,3), (2,4), (4,2), (3,3), (5,4)} When (x,y) is an edge, we say that x is adjacent to y. 1 is adjacent to 2. 2 is not adjacent to 1. 4 is not adjacent to 3.

CS 1034 A “Real-life” Example of a Graph V=set of 6 people: John, Mary, Joe, Helen, Tom, and Paul, of ages 12, 15, 12, 15, 13, and 13, respectively. E ={(x,y) | if x is younger than y} John Joe MaryHelen TomPaul

CS 1035 Intuition Behind Graphs The nodes represent entities (such as people, cities, computers, words, etc.) Edges (x,y) represent relationships between entities x and y, such as: –“x loves y” –“x hates y” –“x is as smart as y” –“x is a sibling of y” –“x is bigger than y” –‘x is faster than y”, …

CS 1036 Directed vs. Undirected Graphs If the directions of the edges matter, then we show the edge directions, and the graph is called a directed graph (or a digraph) The previous two examples are digraphs If the relationships represented by the edges are symmetric (such as (x,y) is edge if and only if x is a sibling of y), then we don’t show the directions of the edges, and the graph is called an undirected graph.

CS 1037 Examples of Undirected Graphs V=set of 6 people: John, Mary, Joe, Helen, Tom, and Paul, where the first 4 are siblings, and the last two are siblings E ={(x,y) | x and y are siblings} John Joe MaryHelen TomPaul

CS 1038 Definition of Some Graph Related Concepts (Paths) A path in a graph G is a sequence of nodes x 1, x 2, …,x k, such that there is an edge from each node the next one in the sequence For example, in the first example graph, the sequence 4, 1, 2, 3 is a path, but the sequence 1, 4, 5 is not a path because (1,4) is not an edge In the “sibling-of” graph, the sequence John, Mary, Joe, Helen is a path, but the sequence Helen, Tom, Paul is not a path

CS 1039 Definition of Some Graph Related Concepts (Cycles) A cycle in a graph G is a path where the last node is the same as the first node. In the “sibling-of” graph, the sequence John, Mary, Joe, Helen, John is a cycle, but the sequence Helen, Tom, Paul, Helen is not a cycle

CS Graph Connectivity An undirected graph is said to be connected if there is a path between every pair of nodes. Otherwise, the graph is disconnected Informally, an undirected graph is connected if it hangs in one piece Disconnected Connected

CS Graph Cyclicity An undirected graph is cyclic if it has at least one cycle. Otherwise, it is acyclic Disconnected and acyclic Connected and acyclic Disconnected and cyclic Connected and cyclic

CS Trees A tree is a connected acyclic undirected graph. The following are three trees:

CS Rooted Trees A rooted tree is a tree where one of the nodes is designated as the root node. (Only one root in a tree) A rooted tree has a hierarchical structure: the root on top, followed by the nodes adjacent to it right below, followed by the nodes adjacent to those next, and so on.

CS Example of a Rooted Tree Unrooted tree Tree rooted with root 1

CS Tree-Related Concepts The nodes adjacent to x and below x are called the children of x, and x is called their parents A node that has no children is called a leaf The descendents of a node are: itself, its children, their children, all the way down The ancestors of a node are: itself, its parent, its grandparent, all the way to the root

CS Tree-Related Concepts (Contd.) The depth of a node is the number of edges from the root to that node. The depth (or height) of a rooted tree is the depth of the lowest leaf Depth of node 10: 3 Depth of this tree:

CS Binary Trees A tree is a binary tree if every node has at most two children Nonbinary treeBinary tree

CS Binary-Tree Related Definitions The children of any node in a binary tree are ordered into a left child and a right child A node can have a left and a right child, a left child only, a right child only, or no children The tree made up of a left child (of a node x) and all its descendents is called the left subtree of x Right subtrees are defined similarly

CS Graphical View Binary-tree Nodes data leftright In practice, a TreeNode will be shown as a circle where the data is put inside, and the node label (if any) is put outside. Graphically, a TreeNode is: 5.82 datalabel A binary-tree node consists of 3 parts: -Data -Pointer to left child -Pointer to right child

CS A Binary-tree Node Class class TreeNode { public: typedef int datatype; TreeNode(datatype x=0, TreeNode *left=NULL, TreeNode *right=NULL){ data=x; this->left=left; this->right=right; }; datatype getData( ) {return data;}; TreeNode *getLeft( ){return left;}; TreeNode *getRight( ){return right;}; void setData(datatype x) {data=x;}; void setLeft(TreeNode *ptr) {left=ptr;}; void setRight(TreeNode *ptr) {right=ptr;}; private: datatype data; // different data type for other apps TreeNode *left; // the pointer to left child TreeNode *right; // the pointer to right child };

CS Binary Tree Class class Tree { public: typedef int datatype; Tree(TreeNode *rootPtr=NULL){this->rootPtr=rootPtr;}; TreeNode *search(datatype x); bool insert(datatype x); TreeNode * remove(datatype x); TreeNode *getRoot(){return rootPtr;}; Tree *getLeftSubtree(); Tree *getRightSubtree(); bool isEmpty(){return rootPtr == NULL;}; private: TreeNode *rootPtr; };

CS Binary Search Trees A binary search tree (BST) is a binary tree where –Every node holds a data value (called key) –For any node x, all the keys in the left subtree of x are ≤ the key of x –For any node x, all the keys in the right subtree of x are > the key of x

CS Example of a BST

CS Searching in a BST To search for a number b: 1.Compare b with the root; –If b=root, return –If b<root, go left –If b>root, go right 2.Repeat step 1, comparing b with the new node we are at. 3.Repeat until either the node is found or we reach a non-existing node Try it with b=12, and also with b=

CS Code for Search in BST // returns a pointer to the TreeNode that contains x, // if one is found. Otherwise, it returns NULL TreeNode * Tree::search(datatype x){ if (isEmpty()){return NULL;} TreeNode *p=rootPtr; while (p != NULL){ datatype a = p->getData(); if (a == x) return p; else if (x getLeft(); else p=p->getRight(); } return NULL; };

CS Insertion into a BST Insert(datatype b, Tree T): 1.Search for the position of b as if it were in the tree. The position is the left or right child of some node x. 2.Create a new node, and assign its address to the appropriate pointer field in x 3.Assign b to the data field of the new node

CS Illustration of Insert Before inserting After inserting 25

CS Code for Insert in BST bool Tree::insert(datatype x){ if (isEmpty()) {rootPtr = new TreeNode(x);return true; } TreeNode *p=rootPtr; while (p != NULL){ datatype a = p->getData(); if (a == x) return false; // data is already there else if (x<a){ if (p->getLeft() == NULL){ // place to insert TreeNode *newNodePtr= new TreeNode(x); p->setLeft(newNodePtr); return true;} elsep=p->getLeft(); }else { // a>a if (p->getRight() == NULL){ // place to insert TreeNode *newNodePtr= new TreeNode(x); p->setRight(newNodePtr); return true;} elsep=p->getRight();} } };

CS Deletion from a BST Illustration in class

CS Deletion from a BST (pseudocode) Delete(datatype b, Tree T) 1.Search for b in tree T. If not found, return. 2.Call x the first node found to contain b 3.If x is a leaf, remove x and set the appropriate pointer in the parent of x to NULL 4.If x has only one child y, remove x, and the parent of x become a direct parent of y (More on the next slide)

CS Deletion (contd.) 5. If x has two children, go to the left subtree, and find there in largest node, and call it y. The node y can be found by tracing the rightmost path until the end. Note that y is either a leaf or has no right child 6. Copy the data field of y onto the data field of x 7. Now delete node y in a manner similar to step 4.

CS Code for Delete in BST(4 slides) // finds x in the tree, removes it, and returns a pointer to the containing // TreeNode. If x is not found, the function returns NULL. TreeNode * Tree::remove(datatype x){ if (isEmpty()) return NULL; TreeNode *p=rootPtr; TreeNode *parent = NULL; // parent of p char whatChild; // 'L' if p is a left child, 'R' O.W. while (p != NULL){ datatype a = p->getData(); if (a == x) break;// x found else if(x getLeft();} else {parent = p; whatChild = 'R'; p=p->getRight();} } if (p==NULL) return NULL; // x was not found

CS // Handle the case where p is a leaf. // Turn the appropriate pointer in its parent to NULL if (p->getLeft() == NULL && p->getRight() == NULL){ if (parent != NULL) // x is not at the root if (whatChild == 'L') parent->setLeft(NULL); else parent->setRight(NULL); else // x is at the root rootPtr=NULL; return p; }

CS else if (p->getLeft() == NULL){ // p has only one a child -- a right child. Let the parent of p // become an immediate parent of the right child of p. if (parent != NULL) // p is not the root if (whatChild == 'L') parent->setLeft(p->getRight()); else parent->setRight(p->getRight()); else rootPtr=p->getRight(); // p is the root return p; } else if (p->getRight() == NULL){ // p has only one a child -- a left child. Let the parent of p // become an immediate parent of the left child of p. if (parent != NULL) // p is not the root if (whatChild == 'L')parent->setLeft(p->getLeft()); elseparent->setRight(p->getLeft()); else rootPtr=p->getLeft(); // p is the root return p; }

CS else { // p has two children TreeNode *returnNode= new TreeNode(*p); // replicates p TreeNode * leftChild = p->getLeft(); if (leftChild->getRight() == NULL){// leftChild has no right child p->setData(leftChild->getData()); p->setLeft(leftChild->getLeft()); delete leftChild; return returnNode; } TreeNode * maxLeft = leftChild->getRight(); TreeNode * parent2 = leftChild; while (maxLeft != NULL){parent2 = maxLeft; maxLeft = maxLeft ->getRight();} // now maxLeft is the node to swap with p. p->setData(maxLeft->getData()); if (maxLeft->getLeft()==NULL) parent2->setRight(NULL); // maxLeft a leaf else parent2->setRight(maxLeft->getLeft()); //maxLeft not a leaf delete maxLeft; return returnNode; } };

CS Additional Things for YOU to Do Add a method to the Tree class for returning the maximum value in the BST Add a method to the Tree class for returning the minimum value in the BST Write a function that takes as input an array of type datatype, and an integer n representing the length of the array, and returns a BST Tree Object containing the elements of the input array