Report

1. 2. 3. 4. General Trees Binary Search Trees AVL Trees Heap Trees Insertion ◦ FIFO ◦ LIFO ◦ Key-sequenced Insertion Deletion Changing a General Tree into a Binary Tree Given the parent Node a new node may be inserted as FIFO First in-first out (FIFO) insertion Data Structures: A Pseudocode Approach with C 5 Insertion Given the parent Node a new node may be inserted as FIFO LIFO Last in-first out (LIFO) insertion Data Structures: A Pseudocode Approach with C 7 Insertion Given the parent Node a new node may be inserted as FIFO LIFO Key-sequenced Insertion Key-sequenced insertion Data Structures: A Pseudocode Approach with C 9 Insertion ◦ FIFO ◦ LIFO ◦ Key-sequenced Insertion Deletion Changing a General Tree into a Binary Tree For general trees nodes to be deleted are restricted to be “leaves” Otherwise a node maybe “purged”, i.e. a node is deleted along with all its children Insertion ◦ FIFO ◦ LIFO ◦ Key-sequenced Insertion Deletion Changing a General Tree into a Binary Tree Changing the meaning of the two pointers: Leftchild …..first child Rightchild ….. Next siblings Changing a General Tree to a Binary Tree Data Structures: A Pseudocode Approach with C 14 Changing a General Tree to a Binary Tree Data Structures: A Pseudocode Approach with C 15 Changing a General Tree to a Binary Tree Data Structures: A Pseudocode Approach with C 16 1. 2. 3. 4. General Trees Binary Search Trees AVL Trees Heap Trees Basic Concepts BST Operations Threaded Trees All items in left subtree < root All items in right subtree > root A binary search tree Not a binary search tree Two binary search trees representing the same set: Basic Concepts BST Operations Threaded Trees Traversal Search ◦ Smallest ……….. ? ◦ Largest …………? ◦ Specific element Insertion Deletion Print out all the keys in sorted order Inorder: 2, 3, 4, 6, 7, 9, 13, 15, 17, 18, 20 Traversal Search ◦ Smallest ……….. ? ◦ Largest …………? ◦ Specific element Insertion Deletion Return the node containing the smallest element in the tree Start at the root and goes 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) 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 BST Traversal Search ◦ Smallest ……….. ? ◦ Largest …………? ◦ Specific element Insertion Deletion 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) Traversal Search ◦ Smallest ……….. ? ◦ Largest …………? ◦ Specific element Insertion 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. Three cases: (1) the node is a leaf ◦ Delete it immediately (2) the node has one sub-tree (right or left) ◦ Adjust a pointer from the parent to bypass that node (3) the node has 2 children ◦ replace the key of that node with the minimum element at the right subtree (or the maximum element at the left subtree) ◦ delete the 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) Basic Concepts BST Operations Threaded Trees Sparing recursion and stack Making use of null right child of leaves to point to next node 1. 2. 3. 4. General Trees Binary Search Trees AVL Trees Heap Trees Properties Operations It is a balanced binary tree (definition of Russian mathematicians Adelson-Velskii and Landis) The height of its sub-trees differs by no more than one (its balance factor is -1, 0, or 1), and its subtrees are also balanced. A sub tree is called Left high (LH) if its balance is 1 Equally high (EH) if it is 0 Right high (RH) if it is -1 Insertion and deletion are same as in BST If unbalance occurs corresponding rotations must be performed to restore balance Steps: ◦ Check if case is case 1 or 2 of the following and act accordingly ◦ Case 1: tree is left high & out-of-balance is created by a adding node to the left of the left sub-tree ◦ …… One right rotation is needed Rotate out-of-balance node right •Case 1 * Tree is left balanced unbalance is caused by node on the left of left sub-tree h+2 h+1 h+1 h+1 h h h h h ◦ Case 2: tree is left high out-of-balance is created by a adding node to the right of the left sub-tree ◦ …… Two rotations are needed: Move from bottom of left sub-tree upwards till an unbalanced node is found and rotate it left Rotate left sub-tree right Add node to right of left balanced subtree h+2 h+1 h+1h h h h+2 h+2 h+1 h+1 h First rotation .. Left rotation of unbalanced node c Second rotation … Right rotation of left sub-tree g h