Trees, Binary Trees, and Binary Search Trees

Trees Linear access time of linked lists is prohibitive Trees Does there exist any simple data structure for which the running time of most operations (search, insert, delete) is O(log N)? Trees Basic concepts Tree traversal Binary tree Binary search tree and its operations

Trees A tree T is a collection of nodes T can be empty (recursive definition) If not empty, a tree T consists of a (distinguished) node r (the root), and zero or more nonempty sub-trees T1, T2, ...., Tk

Tree can be viewed as a ‘nested’ lists: Tree is also a graph … list of lists of lists … Tree is also a graph …

Some Terminologies Child and Parent Leaves Sibling Every node except the root has one parent  A node can have a zero or more children Leaves Leaves are nodes with no children Sibling nodes with same parent

More Terminologies Path Length of a path Depth of a node A sequence of edges Length of a path number of edges on the path Depth of a node length of the unique path from the root to that node Height of a node length of the longest path from that node to a leaf all leaves are at height 0 The height of a tree = the height of the root = the depth of the deepest leaf Ancestor and descendant If there is a path from n1 to n2 n1 is an ancestor of n2, n2 is a descendant of n1 Proper ancestor and proper descendant

Example: UNIX Directory

Tree Operations Traversal, the most important we will not implement a general ‘tree’, so wont discuss Search Insertion Deletion

Tree Traversal Used to print out the data in a tree in a certain order Pre-order traversal Print the data at the root Recursively print out all data in the leftmost subtree … Recursively print out all data in the rightmost subtree

A drawing of tree with two pointers … Struct TreeNode { double element; // the data TreeNode* child; // FIRST child : go to the next generation TreeNode* next; // next SIBLING : go to the same generation }

preorder(tree) if empty(tree) stop else print(element(tree)) preorder(child(tree)) preorder(next(tree))

Example: Unix Directory Traversal PreOrder PostOrder

Binary Trees A tree in which no node can have more than two children The depth of an “average” binary tree is considerably smaller than N, even though in the worst case, the depth can be as large as N – 1. Generic binary tree Worst-case binary tree

Binary Tree ADT Implementation at most two children, we can keep direct pointers to them a linked list is physically a pointer, so is a tree! Possible operations on the Binary Tree ADT Parent, left_child, right_child, sibling, root, etc Tree operations: Search, insertion and deletion Make a Binary Tree ADT later …

A drawing of linked list with one pointer … A drawing of binary tree with two pointers … Struct BinaryNode { double element; // the data BinaryNode* left; // left child BinaryNode* right; // right child }

Example: Expression Trees Leaves are operands (constants or variables) The internal nodes contain operators Will not be a binary tree if some operators are not binary

Preorder, Postorder and Inorder Preorder traversal node, left, right prefix expression ++a*bc*+*defg

Preorder, Postorder and Inorder Inorder traversal left, node, right infix expression a+b*c+d*e+f*g Postorder traversal left, right, node postfix expression abc*+de*f+g*+

Preorder, Postorder and Inorder Pseudo Code

Binary tree insertion and deletion …

BST: Binary Search Tree

Recall the binary search algorithm   4 7 10 15 19 20 42 54 87 90

Binary Search Trees (BST) A data structure for efficient searching, inser-tion and deletion Binary search tree property For every node X All the keys in its left subtree are smaller than the key value in X All the keys in its right subtree are larger than the key value in X

Binary Search Trees A binary search tree Not a binary search tree

Binary Search Trees Average depth of a node is O(log N) Maximum depth of a node is O(N) The same set of keys may have different BSTs

Searching BST If we are searching for 15, then we are done. If we are searching for a key < 15, then we should search in the left subtree. If we are searching for a key > 15, then we should search in the right subtree.

Searching (Find) Find X: return a pointer to the node that has key X, or NULL if there is no such node Time complexity: O(height of the tree) find(const double x, BinaryNode* t) const

Inorder Traversal of BST Inorder traversal of BST prints out all the keys in sorted order Inorder: 2, 3, 4, 6, 7, 9, 13, 15, 17, 18, 20

findMin/ findMax Goal: return the node containing the smallest (largest) key in the tree Algorithm: Start at the root and go left (right) as long as there is a left (right) child. The stopping point is the smallest (largest) element Time complexity = O(height of the tree) BinaryNode* findMin(BinaryNode* t) const

Insertion Proceed down the tree as you would with a find If X is found, do nothing (or update something) Otherwise, insert X at the last spot on the path traversed Time complexity = O(height of the tree)

void insert(double x, BinaryNode*& t) { if (t==NULL) t = new BinaryNode(x,NULL,NULL); else if (x<t->element) insert(x,t->left); else if (t->element<x) insert(x,t->right); else ; // do nothing }

Insertion Example Construct a BST successively from a sequence of data: 35,60,2,80,40,85,32,33,31,5,30

Deletion When we delete a node, we need to consider how we take care of the children of the deleted node. This has to be done such that the property of the search tree is maintained.

Deletion under Different Cases Case 1: the node is a leaf Delete it immediately Case 2: the node has one child Adjust a pointer from the parent to bypass that node

Deletion Case 3 Case 3: the node has 2 children Replace the key of that node with the minimum element at the right subtree (or replace the key by the maximum at the left subtree!) Delete that minimum element Has either no child or only right child because if it has a left child, that left child would be smaller and would have been chosen. So invoke case 1 or 2. Time complexity = O(height of the tree)

void remove(double x, BinaryNode*& t) { if (t==NULL) return; if (x<t->element) remove(x,t->left); else if (t->element < x) remove (x, t->right); else if (t->left != NULL && t->right != NULL) // two children t->element = finMin(t->right) ->element; remove(t->element,t->right); } else Binarynode* oldNode = t; t = (t->left != NULL) ? t->left : t->right; delete oldNode;

Deletion Example Removing 40 from (a) results in (b) using the smallest element in the right subtree (i.e. the successor) 5 30 2 40 80 35 32 33 31 85 60 5 30 2 60 80 35 32 33 31 85 (a) (b)

Removing 40 from (a) results in (c) using the largest element in the left subtree (i.e., the predecessor) 5 30 2 40 80 35 32 33 31 85 60 5 30 2 35 80 32 33 31 85 60 (a) (c) 39

Removing 30 from (c), we may replace the element with either 5 (predecessor) or 31 (successor). If we choose 5, then (d) results. 5 30 2 35 80 32 33 31 85 60 2 5 35 80 32 33 31 85 60 (c) (d) 40

Example: compute the number of nodes?

Example: Successor The successor of a node x is defined as: The node y, whose key(y) is the successor of key(x) in sorted order sorted order of this tree. (2,3,4,6,7,9,13,15,17,18,20) Some examples: Which node is the successor of 2? Which node is the successor of 9? Which node is the successor of 13? Which node is the successor of 20? Null Search trees 42

Scenario I: Node x Has a Right Subtree By definition of BST, all items greater than x are in this right sub-tree. Successor is the minimum( right( x ) ) maybe null Search trees 43

Scenario II: Node x Has No Right Subtree and x is the Left Child of Parent (x) Successor is parent( x ) Why? The successor is the node whose key would appear in the next sorted order. Think about traversal in-order. Who would be the successor of x? The parent of x! Search trees 44

Scenario III: Node x Has No Right Subtree and Is Not a Left-Child of an Immediate Parent Keep moving up the tree until you find a parent which branches from the left(). Successor of x y Stated in Pseudo code. x Search trees 45

Three Scenarios to Determine Successor Successor(x) x has right descendants => minimum( right(x) ) x has no right descendants Scenario I x is the left child of some node => parent(x) x is the right child of some node Scenario II Scenario III Search trees 46

Successor Pseudo-Codes Verify this code with this tree. Find successor of 3  4 9  13 13  15 18  20 Note that parent( root ) = NULL Scenario I Scenario II Scenario III Search trees 47

Problem If we use a “doubly linked” tree, finding parent is easy. But usually, we implement the tree using only pointers to the left and right node.  So, finding the parent is tricky. For this implementation we need to store the ‘path’  Stack! class Node { int data; Node * left; Node * right; Node * parent; }; class Node { int data; Node * left; Node * right; }; 48

Use a Stack to Find Successor PART I Initialize an empty Stack s. Start at the root node, and traverse the tree until we find the node x. Push all visited nodes onto the stack. PART II Once node x is found, find successor using 3 scenarios mentioned before. Parent nodes are found by popping the stack! 49

Example Successor(root, 13) Part I Traverse tree from root to find 13 order -> 15, 6, 7, 13 7 6 15 Stack s 50

7 6 15 Stack s Successor(root, 13) Part II Find Parent (Scenario III) y=s.pop() while y!=NULL and x=right(y) x = y; if s.isempty() y=NULL else return y y =pop()=6 y =pop()=7 x = 13 7 6 15 Stack s 51

Make a binary or BST ADT …

For a generic (binary) tree: Struct Node { double element; // the data Node* left; // left child Node* right; // right child } class Tree { public: Tree(); // constructor Tree(const Tree& t); ~Tree(); // destructor bool empty() const; double root(); // decomposition (access functions) Tree& left(); Tree& right(); bool search(const double x); void insert(const double x); // compose x into a tree void remove(const double x); // decompose x from a tree private: Node* root; Composition of a tree is not completely given by the constructors, The composition is a level above ‘constructor’ (insert and remove are different from those of BST)

For BST tree: Struct Node { double element; // the data Node* left; // left child Node* right; // right child } class BST { public: BST(); // constructor BST(const Tree& t); ~BST(); // destructor bool empty() const; double root(); // decomposition (access functions) BST& left(); BST& right(); bool serch(const double x); // search an element void insert(const double x); // compose x into a tree void remove(const double x); // decompose x from a tree private: Node* root; Composition of a tree is actually given by the constructors already BST is for efficient search, insertion and removal, so restricting these functions.

Weiss textbook: class BST { public: BST(); BST(const Tree& t); ~BST(); bool empty() const; bool search(const double x); // contains void insert(const double x); // compose x into a tree void remove(const double x); // decompose x from a tree private: Struct Node { double element; Node* left; Node* right; Node(…) {…}; // constructuro for Node } Node* root; void insert(const double x, Node*& t) const; // recursive function void remove(…) Node* findMin(Node* t); void makeEmpty(Node*& t); // recursive ‘destructor’ bool contains(const double x, Node* t) const; Weiss textbook: Trees with different implemntations are called differently, For example, a heap is usually implemneted with an array

Comments: root, left subtree, right subtree are missing: 1. we can’t write other tree algorithms, is implementation dependent, BUT, 2. this is only for BST (we only need search, insert and remove, may not need other tree algorithms) so it’s two layers, the public for BST, and the private for Binary Tree. 3. it might be defined internally in ‘private’ part (actually it’s implicitly done). Trees with different implemntations are called differently, For example, a heap is usually implemneted with an array

A public non-recursive member function: void insert(double x) { insert(x,root); } A private recursive member function: void insert(double x, BinaryNode*& t) { if (t==NULL) t = new BinaryNode(x,NULL,NULL); else if (x<t->element) insert(x,t->left); else if (t->element<x) insert(x,t->right); else ; // do nothing }

By inheritance All search, insert and deletion have to be redefined. Struct Node { double element; // the data Node* left; // left child Node* right; // right child } Class BinaryTree { … class BST : public BinaryTree { void BST::search () { void BST::insert () { void BST::delete () { template<typename T> Struct Node { T element; // the data Node* left; // left child Node* right; // right child } class BinaryTree { … class BST : public BinaryTree<T> { void BST<T>::search (const T& x) { void BST<T>::insert (…) { void BST<T>::delete (…) { Composition of a tree is actually given by the constructors already All search, insert and deletion have to be redefined.

More general BST … template<typename T> Struct Node { T element; // the data Node* left; // left child Node* right; // right child } class BinaryTree { … class BST : public BinaryTree<T> { void BST<T>::search (const T& x) { void BST<T>::insert (…) { void BST<T>::delete (…) { template<typename T> class BinaryTree { … } template<typename T, typename K> class BST : public BinaryTree<T> { void BST<T>::search (const K& key) { Composition of a tree is actually given by the constructors already Search ‘key’ of K might be different from the data record of T!!!

Deletion Code (1/4) First Element Search, and then Convert Case III, if any, to Case I or II template<class E, class K> BSTree<E,K>& BSTree<E,K>::Delete(const K& k, E& e) { // Delete element with key k and put it in e. // set p to point to node with key k (to be deleted) BinaryTreeNode<E> *p = root, // search pointer *pp = 0; // parent of p while (p && p->data != k){ // move to a child of p pp = p; if (k < p->data) p = p->LeftChild; else p = p->RightChild; } 60

Deletion Code (2/4) if (!p) throw BadInput(); // no element with key k e = p->data; // save element to delete // restructure tree // handle case when p has two children if (p->LeftChild && p->RightChild) { // two children convert to zero or one child case // find predecessor, i.e., the largest element in // left subtree of p BinaryTreeNode<E> *s = p->LeftChild, *ps = p; // parent of s while (s->RightChild) { // move to larger element ps = s; s = s->RightChild; } 61

Deletion Code (3/4) // move from s to p p->data = s->data; p = s; // move/reposition pointers for deletion pp = ps; } // p now has at most one child // save child pointer to c for adoption BinaryTreeNode<E> *c; if (p->LeftChild) c = p->LeftChild; else c = p->RightChild; // deleting p if (p == root) root = c; // a special case: delete root else { // is p left or right child of pp? if (p == pp->LeftChild) pp->LeftChild = c;//adoption else pp->RightChild = c; 62

Deletion Code (4/4) delete p; return *this; } 63