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1. General Trees 2. Binary Search Trees 3. AVL Trees 4. Heap Trees

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Insertion ◦ FIFO ◦ LIFO ◦ Key-sequenced Insertion Deletion Changing a General Tree into a Binary Tree

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Given the parent Node a new node may be inserted as FIFO

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Data Structures: A Pseudocode Approach with C 5 First in-first out (FIFO) insertion

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Given the parent Node a new node may be inserted as FIFO LIFO Insertion

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Data Structures: A Pseudocode Approach with C 7 Last in-first out (LIFO) insertion

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Given the parent Node a new node may be inserted as FIFO LIFO Key-sequenced Insertion Insertion

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Data Structures: A Pseudocode Approach with C 9 Key-sequenced insertion

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Insertion ◦ FIFO ◦ LIFO ◦ Key-sequenced Insertion Deletion Changing a General Tree into a Binary Tree

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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

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Insertion ◦ FIFO ◦ LIFO ◦ Key-sequenced Insertion Deletion Changing a General Tree into a Binary Tree

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Changing the meaning of the two pointers: Leftchild …..first child Rightchild ….. Next siblings

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Data Structures: A Pseudocode Approach with C 14 Changing a General Tree to a Binary Tree

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Data Structures: A Pseudocode Approach with C 15 Changing a General Tree to a Binary Tree

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Data Structures: A Pseudocode Approach with C 16 Changing a General Tree to a Binary Tree

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1. General Trees 2. Binary Search Trees 3. AVL Trees 4. Heap Trees

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Basic Concepts BST Operations Threaded Trees

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All items in left subtree < root All items in right subtree > root

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A binary search tree Not a binary search tree

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Two binary search trees representing the same set:

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Basic Concepts BST Operations Threaded Trees

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Traversal Search ◦ Smallest ……….. ? ◦ Largest …………? ◦ Specific element Insertion Deletion

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Print out all the keys in sorted order Inorder: 2, 3, 4, 6, 7, 9, 13, 15, 17, 18, 20

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Traversal Search ◦ Smallest ……….. ? ◦ Largest …………? ◦ Specific element Insertion Deletion

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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)

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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.

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Searching BST

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Traversal Search ◦ Smallest ……….. ? ◦ Largest …………? ◦ Specific element Insertion Deletion

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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)

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Traversal Search ◦ Smallest ……….. ? ◦ Largest …………? ◦ Specific element Insertion Deletion

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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.

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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

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(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)

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Basic Concepts BST Operations Threaded Trees

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Sparing recursion and stack Making use of null right child of leaves to point to next node

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1. General Trees 2. Binary Search Trees 3. AVL Trees 4. Heap Trees

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Properties Operations

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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.

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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

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Insertion and deletion are same as in BST If unbalance occurs corresponding rotations must be performed to restore balance

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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

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h h hh h h+1h+2 h+1 Case 1 * Tree is left balanced unbalance is caused by node on the left of left sub-tree

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◦ 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

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h+2h+1 Add node to right of left balanced subtree First rotation.. Left rotation of unbalanced node c Second rotation … Right rotation of left sub-tree g h h hh h h+1 h+2 h+1

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