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Binary Search Trees Azhar Maqsood School of Electrical Engineering and Computer Sciences (SEECS-NUST)

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When we store ordered data in an array, we have a very efficient search algorithm, the binary search. However, we have very inefficient insertion and deletion algorithms that require shifting data in the array.

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To provide for efficient insertions and deletions, we developed the linked list. The problem with linked lists, however, is that their search algorithms are sequential searches, which are very inefficient. Can’t we get the best of both worlds (i.e., efficient search, insertion, and deletion algorithms)?

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The binary search tree is our answer! The binary search tree is a binary tree with the following properties: ◦ All items (keys) in the left subtree are less than the root’s key. ◦ All items (keys) in the right subtree are greater than the root’s key. ◦ Each subtree is, itself, a binary search tree.

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◦ Assume each node of a binary tree stores a data item ◦ Assume data items are of some type that be ordered and all items are distinct. No two items have the same value. ◦ A binary search tree is a binary tree such that ◦ for every node X in the tree: the values of all the items in its left subtree are smaller than the value of the item in X the values of all items in its right subtree are greater than the value of the item in X.

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Here you can see the basic structure of a binary search tree.

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A binary search tree is organized as the name suggests A tree can be represented by a linked data structure where each node is an object In addition to a key field, each node contains fields left, and right, that correspond to its left child and right child LeftRightKey LeftRightKeyLeftRightKey LeftRightKey

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If a child or parent is missing, the appropriate field contains the value NULL. The root is only node in the tree, whose parent field is NULL. Top of the tree is known as ‘root’ and the exposed nodes at bottom are known as ‘leafs’.

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6 2 8 14 3 6 2 8 14 37 A binary search tree Not a binary search tree, but a binary tree

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- a & b are full and BST -a, b, & d are complete -c and e are skewed

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Here we see examples of binary trees that are not binary search trees … why? a) 22 > 17 b) 11 < 17 c) 11 > 6 d) 22 > 17

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Note: we have written this definition in a way that ensures that no two entries in a binary search tree can have equal keys. Although it is possible to modify the definition to allow entries with duplicate keys, it makes the algorithms somewhat more complicated. If duplicates need to be accounted for, they can be stored in another data structure (e.g., list).

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The search tree property does not say what happens with the elements with the same key. In our examples, all the keys will be distinct but in practice it is important to cover the case of multiple elements with the same key (duplicate keys). The duplicate keys can all be kept in the left subtree, or all in the right subtree. It doesn’t matter which we choose, but it matters that the choice is the same for the whole implementation. Another issue: with duplicate keys, it is important to have extra operations in the interface: getAll, and removeAll

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Binary trees offer short paths from root A node has up to two children Data is organized by value Operations: ◦ Insertion ◦ Search ◦ Traversal ◦ Deletion

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Find Minimum Find the item that has the minimum value in the tree Find Maximum Find the item that has the maximum value in the tree Print Print the values of all items in the tree using a traversal strategy that is appropriate for the application Successor Predecessor

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The first value inserted goes at the root Every node inserted becomes a leaf Descend left or right depending on value

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6 28 NULL 1 4 3 t Tree root node 5 NULL New Node t t Inserting Item 5 to the Tree

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The BST property allows us to print out all keys in BST in sorted order using a simple recursive algorithm. InOrder_Tree_Walk(x) { if x≠ NULL InOrder_Tree_Walk(left[x]) print key[x] InOrder_Tree_Walk(right[x]) } InOrder_Tree_Walk(x) { if x≠ NULL InOrder_Tree_Walk(left[x]) print key[x] InOrder_Tree_Walk(right[x]) }

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72038 4, 7, 16, 20, 37, 38, 43

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BST-inorder-traverse (node) 1. If node has left child BST-inorder-traverse (Left(node)) 2. Apply operation to node (e.g. cout) 3. If node has right child BST-inorder-traverse (Right(node)) Order for tree above: 1, 2, 3, 5, 6, 7 Application: Display tree in ascending order

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BST-preorder-traverse (node) 1. Apply operation to node e.g cout 2. If node has left child BST-preorder-traverse (Left(node)) 3. If node has right child BST-preorder-traverse (Right(node)) Order for tree above: 3, 1, 2, 6, 5, 7 Application: Copy BST

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BST-postorder-traverse (node) 1. If node has left child BST-postorder-traverse (Left(node)) 2. If node has right child BST-postorder-traverse (Right(node)) 3. Apply operation to node e.g cout Order for tree above: 2, 1, 5, 7, 6, 3 Application: Deallocate BST

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Want to search for key stored in BST Functions: Tree search Minimum Maximum Successor Predecessor

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Start at the root Is target = value at current node? ◦ We’re done Is target < value at current node? ◦ Yes: search left subtree ◦ No: search right subtree

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Search Starts from the Root

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When given a pointer to the root of a tree and a key, Tree-Search will return a pointer to the node with key k if one exists (x = root). Tree-Search (x, k) { if x=NULL or k=key[ x] return x if k

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This function follows the BST property, if value of k is smaller than that of x, then it should be in left sub tree of x, else it should be in right sub tree of x. 6 2 8 14 3

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The minimum element of BST is the left most node of left sub tree. Therefore the minimum can always be found by tracking the left child pointers until an empty sub tree is reached. If there is no left sub tree then minimum key is at root( i.e. key[ x]) Tree-Minimum(x) { while( left[ x] != NULL) x = left [x] return x }

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The maximum element of BST is the right most node of right sub tree. Therefore the maximum can always be found by tracking the right child pointers until an empty sub tree is reached. Tree-Maximum( x) { while( right[ x] != NULL) x = right [x] return x }

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The successor of a node x, is node y, that has the smallest key greater than that of x. ◦ If x has a right subtree, then successor(x) is the left most element in that sub tree. ◦ If x has no right sub tree, then successor(x) is the lowest ancestor of x (above x on the path to the root) that has x in its left sub tree.

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The successor of node with key 15 is node with key 17. The successor of node with key 7 is node with key 9. The successor of node with key 13 is node with key 15.

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The predecessor is the node that has the largest key smaller than that of x ◦ If x has a left sub tree, then the predecessor must be the right most element of the left sub tree. ◦ If x has no left subtree, then predecessor (x) is the lowest ancestor of x (above x on the path to the root) that has x in its right subtree.

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The predecessor of node with key 6 is node with key 4. The predecessor of node with key 15 is node with key 13. The predecessor of node with key 17 is node with key 15.

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If z has no children, we modify its parent to replace z with NULL as child. If z has one child, we splice out z by making a new link between its child and its parent. If node has two children, we splice out z’s successor y, which has no left child and replace the contents of z with the contents of y.

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Three cases for remove(x): x occurs at a leaf: simply remove the node containing x 25 15 5 20 18 40 45 removeElement(45) -------------> 25 15 5 20 18 40

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x occurs at a node with one child v : remove the node containing x and make v a direct child of x ’s parent (if any). 25 15 5 20 18 40 45 removeElement(20) -------------> 25 15 5 18 40 45

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x occurs at a node with two children: first replace x with smallest value in right subtree of x. This value occurs at a node with no left child. So we can delete this node using one of the two previous cases. 25 15 5 20 18 40 45 removeElement(15) -------------> 25 18 5 20 18 40 45 Now remove this<---

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6 2 8 15 3 4 6 3 8 15 3 4 Deletion of node 2. before after

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template void BinarySearchTree :: remove( const Comparable & x, BinaryNode * & t ) const { if( t == NULL ) return; // Item not found; do nothing if( x 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 = findMin( t->right )->element; remove( t->element, t->right ); } else { BinaryNode *oldNode = t; t = ( t->left != NULL ) ? t->left : t->right; delete oldNode; }

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