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This comprehensive overview covers linear search, binary search, binary search trees, algorithms, time complexity, and insertion techniques in depth for academic purpose.
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Searching andBinary Search Trees CSCI 3333 Data Structures
Acknowledgement • Dr. Yue • Mr. Charles Moen • Dr. Wei Ding • Ms. Krishani Abeysekera • Dr. Michael Goodrich • Dr. Richard F. Gilberg
Searching • Google: era of searching! • Many type of searching: • Search engines: best match. • Key searching: searching for records with specified key values. • Keys are usually assumed to come from a total order (e.g. integers or strings) • Keys are usually assumed to be unique. No two records have the same key.
Naïve Linear Search Algorithm LinearSearch(r, key) Input: r: an array of record. Each record contains a unique key. key: key to be found. Output: i: the index such that r[i] contains the key, or -1 if not found. foreach element in r with index i if r[i].key = key return i; end foreach return -1
Naïve Linear Search • Linear time in finding! • Only suitable when n is very small. • No ‘pre-processing’.
Binary Search • Assume the array has already been sorted in ascending key value. • Example: find(7) 0 1 3 4 5 7 8 9 11 14 16 18 19 m h l 0 1 3 4 5 7 8 9 11 14 16 18 19 m h l 0 1 3 4 5 7 8 9 11 14 16 18 19 m h l 0 1 3 4 5 7 8 9 11 14 16 18 19 l=m =h
Binary Search Algorithm Algorithm BinarySearch(r, key) Input: r: an array of record sorted by its key values. Each record contains a unique key. key: key to be found. Output: i: the index such that r[i] contains the key, or -1 if not found. low = 0; high = r.size – 1 while (low <= high) do mid = (high – low) / 2 if r[mid].key > key then high = mid-1 if r[mid].key = key then return mid; if r[mid].key < key then low = mid + 1 end while return -1;
Binary Search • O(lg n) in finding! • Especially suitable for relatively static sets of records. • Question: can we find a searching algorithm of O(1)? • Question: can we find a searching algorithm of O(lg N), where the time complexity for insert and delete is better? • Average case • Worst case
Binary Search Trees • A binary search tree (BST) is a binary tree which has the following properties: • Each node has a unique key value. • The left subtree of a node contains only values less than the node's value. • The right subtree of a node contains only key values greater than the node's value. • Each subtree is itself a binary search tree.
BST • An inorder traversal of a binary search trees visits the keys in increasing order. • Note that there are variations in the formulation of BST. The one defined in the textbook is somewhat different and is best mapped to implement a dictionary. • Duplicate key values allowed. • Records stored in leaf nodes. • Leaf nodes store no key. • The version of BST used here is more basic, easier to understand, and more popular.
Searching a BST • Similar to binary search. • Each key comparison with a node will result in one of these: • target < node key => search left subtree. • target = node key => found! • target > node key => search right subtree. • Difference with binary search: do not always eliminate about half of the candidates. • The BST may not be balanced.
BST Find Algorithm (Recursive) Algorithm Find(r, target) Input: r: root of a BST. target: key to be found. Output: p: the node in the BST that contains the target; null if not found. if r = null return null; if target = r.key return r; if target < r.key return find(r.leftChild(),target) else // target > r.key return find(r.rightChild(),target) end if
BST Find Algorithm (Iterative) Algorithm Find(r, target) Input: r: root of a BST. target: key to be found. Output: p: the node in the BST that contains the target; null if not found. result = r while result != null do if target = result.key return result; if target < result.key then result = result.leftChild() else // target > r.key result = result.rightChild() end if end while return result
Time Complexity • Depend on the shape of the BST. • Worst case: O(n) for unbalanced trees. • Average case: O(lg N)
BST Insert • Insert a record rec with key rec.key. • Idea: search for rec.key in the BST. • If found => error (keys are supposed to be unique) • If not found => ready to insert (however, need to remember the parent node for insertion).
BST Insert Algorithm (Iterative) Algorithm Insert(r, rec) Input: r: root of a BST. rec: record to be found. Output: whether the insertion is successful. Side Effect: r may be updated. curr = r parent = null // find the location for insertion. while (curr != null) do if curr.key = rec.key then return false // duplicate key: unsuccessful insertion else if rec.key < curr.key then curr = curr.leftChild() parent = curr else // rec.key > curr.key curr = curr.rightChild() parent = curr end if end while
BST Insert (Continue) // create new node for inserrtion newNode = new Node(rec) newNode.parent = parent // Insertion. if parent = null then r = newNode; else if rec.key < parent.key then parent.leftChild = newNode else parent.rightChild = newNode end if return true
BST Delete • More complicated • Cannot delete a parent without arranging for the children! • Who take care of the children? Grandparent. • If there is no grandparent? The root is deleted and a child will become the root. • Need to keep track of the parent of the node to be deleted.
BST Delete • Find the node with the key to be deleted. • If not found, deletion is not successful. • Deletion of the node p: • No child: simple deletion • Single child: connect the parent of p to the single child of p. • Both children: • Connect the parent of p to the right child of p. • Connect the left child of p as the left child of the immediate successor of p.
BST Delete Algorithm Algorithm Delete(r, target) Input: r: root of a BST. target: key of the record to be found. Output: whether the deletion is successful. Side Effect: r is updated if the root node is deleted. curr = r; parent = null // find the location for insertion. while (curr != null and curr.key != target) do if target < curr.key then curr = curr.left parent = curr else // target > curr.key curr = curr.right parent = curr end if end while
BST Delete Algorithm (Continue) if curr = null then return false; // not found end if if curr.right = null then // no right child if parent = null then // root deleted r = r.left else if target < parent.key then parent.left = curr.left else parent.right = curr.left end if end if else
BST Delete Algorithm (Continue) // has right child if curr.left = null then // has a right child but no left child if parent = null then // root deleted r = r.right else if target < parent.key then parent.left = curr.right else parent.right = curr.right end if end if else
BST Delete Algorithm (Continue) // has both left and right children // find immediate successor. immedSucc = curr.right while (immedSucc.left != null) immedSucc = immedSucc.left end while immedSucc.left = curr.left
BST Delete Algorithm (Continue) if parent = null then // root deleted r = r.right else if target < parent.key then parent.left = curr.right else parent.right = curr.right end if end if end if delete curr end if
Time Complexity • Worst case: O(N) • Result in unbalanced tree.
Unbalanced BST • Two approaches: • Reorganization when the trees become very unbalanced. • Use balanced BST • Balanced BST • Many potential definitions. • Height = O(lg n)