Download
search binary search trees n.
Skip this Video
Loading SlideShow in 5 Seconds..
Search: Binary Search Trees PowerPoint Presentation
Download Presentation
Search: Binary Search Trees

Search: Binary Search Trees

3 Vues Download Presentation
Télécharger la présentation

Search: Binary Search Trees

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Search: Binary Search Trees Dr. Yingwu Zhu

  2. Review: Linear Search • Collection of data items to be searched is organized in a listx1, x2, … xn • Assume == and < operators defined for the type • Linear search begins with item 1 • continue through the list until target found • or reach end of list

  3. Linear Search Array based search function void LinearSearch (int v[], int n, const int& item, boolean &found, int &loc) { found = false; loc = 0; for ( ; ; ){ if (found || loc >= n) return; if (item == v[loc]) found = true; else loc++; } }

  4. Linear Search Singly-linked list based search function void LinearSearch (NodePointer first, const int& item, bool &found, int &loc) { NodePointer locptr=first; found = false; for{loc=0; !found && locptr!=NULL; locptr=locptr->next){ if (item == locptr->data) found = true; else loc++; } }

  5. Binary Search • Two requirements?

  6. Binary Search • Two requirements • The data items are in ascending order (can they be in decreasing order? ) • Direct access of each data item for efficiency (why linked-list is not good!)

  7. Binary Search Binary search function for vector void LinearSearch (int v[], int n, const t &item, bool &found, int &loc) { found = false; loc = 0; int first = 0; last = n - 1; for ( ; ; ){ if (found || first > last) return; loc = (first + last) / 2; if (item < v[loc]) last = loc - 1; else if (item > v[loc]) first = loc + 1; else found = true; // item found } }

  8. Binary Search vs. Linear Search • Usually outperforms Linear search: O(logn) vs. O(n) • Disadvantages • Sorted list of data items • Direct access of storage structure, not good for linked-list • Good news: It is possible to use a linked structure which can be searched in a binary-like manner

  9. 13 28 35 49 62 66 80 Binary Search Tree (BST) • Consider the following ordered list of integers • Examine middle element • Examine left, right sublist (maintain pointers) • (Recursively) examine left, right sublists

  10. Binary Search Tree • Redraw the previous structure so that it has a treelike shape – a binary tree

  11. Trees • A data structure which consists of • a finite set of elements called nodes or vertices • a finite set of directed arcs which connect the nodes • If the tree is nonempty • one of the nodes (the root) has no incoming arc • every other node can be reached by following a unique sequence of consecutive arcs (or paths)

  12. Root node • Children of the parent (3) • Siblings to each other Leaf nodes Trees • Tree terminology

  13. Binary Trees • Each node has at most two children • Could be empty

  14. Array Representation of Binary Trees • Store the ith node in the ith location of the array

  15. Array Representation of Binary Trees • Works OK for complete trees, not for sparse trees

  16. Some Tree Definition, p656 • Complete trees • Each level is completely filled except the bottom level • The leftmost positions are filled at the bottom level • Array storage is perfect for them • Balanced trees • Binary trees • |left_subtree – right_subtree|<=1 • Tree Height/Depth: • Number of levels

  17. Tree Questions • A complete tree must be a balanced tree? • Give a node with position i in a complete tree, what are the positions of its child nodes? • Left child? • Right child?

  18. Linked Representation of Binary Trees • Uses space more efficiently • Provides additional flexibility • Each node has two links • one to the left child of the node • one to the right child of the node • if no child node exists for a node, the link is set to NULL

  19. Linked Representation of Binary Trees • Example

  20. Inductive step Binary Trees as Recursive Data Structures • A binary tree is either empty … or • Consists of • a node called the root • root has pointers to two disjoint binary (sub)trees called … • right (sub)tree • left (sub)tree Anchor Which is either empty … or … Which is either empty … or …

  21. The "anchor" The inductive step Tree Traversal is Recursive If the binary tree is empty thendo nothing Else N: Visit the root, process dataL: Traverse the left subtreeR: Traverse the right subtree

  22. Traversal Order Three possibilities for inductive step … • Left subtree, Node, Right subtreethe inorder traversal • Node, Left subtree, Right subtreethe preorder traversal • Left subtree, Right subtree, Nodethe postorder traversal

  23. ADT: Binary Search Tree (BST) • Collection of Data Elements • Binary tree • BST property: for each node x, • value in left child of x< value in x < in right child of x • Basic operations • Construct an empty BST • Determine if BST is empty • Search BST for given item

  24. ADT Binary Search Tree (BST) • Basic operations (ctd) • Insert a new item in the BST • Maintain the BST property • Delete an item from the BST • Maintain the BST property • Traverse the BST • Visit each node exactly once • The inordertraversal must visit the values in the nodes in ascending order

  25. BST typedef int T; class BST {public: BST() : myRoot(0) {} bool empty() { return myRoot==NULL; }private: class BinNode { public: T data; BinNode* left, right; BinNode() : left(0), right(0) {} BinNode(T item): data(item), left(0), right(0) {} }; //end of BinNode typedef BinNode* BinNodePtr; BinNodePtr myRoot; };

  26. BST operations • Inorder, preorder and postorder traversals (recursive) • Non-recursive traversals

  27. BST Traversals, p.670 • Why do we need two functions? • Public: void inorder(ostream& out) • Private: inorderAux(…) • Do the actual job • To the left subtree and then right subtree • Can we just use one function? • Probably NO

  28. Non-recursive traversals • Let’s try non-recusive inorder • Any idea to implement non-recursive traversals? What ADT can be used as an aid tool?

  29. Non-recursive traversals • Basic idea • Use LIFO data structure: stack • #include <stack> (provided by STL) • Chapter 9, google stack in C++ for more details, members and functions • Useful in your advanced programming

  30. Non-recursive inorder • Let’s see how it works given a tree • Step by step

  31. Non-recursive inorder 1. Set current node pointer ptr = myRoot 2. Push current node pointer ptr and all the leftmost nodes on its subtree into a stack until meeting a NULL pointer 3. Check if the stack is empty or not. If yes, goto 5, otherwise goto 4 4. Take the top element in stack and assign it to ptr, print our ptr->data, pop the top element, and set ptr = ptr->right (visit right subtree), goto 2 5. Done

  32. Non-recursive inorder void BST::non_recur_inorder(ostream& out) { if (!myRoot) return; // empty BST BinNodePointer ptr = myRoot; stack<BST::BinNodePointer> s; // empty stack for ( ; ; ) { while (ptr) { s.push(ptr); ptr = ptr->left; } // leftmost nodes if (s.empty()) break; // done ptr = s.top(); s.pop(); out << ptr->data << “ “; ptr = ptr->right; // right substree } }

  33. Non-recursive traversals • Can you do it yourself? • Preorder? • Postorder?

  34. BST Searches • Search begins at root • If that is desired item, done • If item is less, move downleft subtree • If item searched for is greater, move down right subtree • If item is not found, we will run into an empty subtree

  35. BST Searches • Write recursive search • Write Non-recursive search algorithm • bool search(const T& item) const

  36. Insertion Operation Basic idea: • Use search operation to locate the insertion position or already existing item • Use a parent point pointing to the parent of the node currently being examined as descending the tree

  37. Insertion Operation • Non-recursive insertion • Recursive insertion • Go through insertion illustration p.677 • Initially parent = NULL, locptr = root; • Location termination at NULL pointer or encountering the item • Parent->left/right = item

  38. Recursive Insertion • See p. 681 • Thinking! • Why using & at subtreeroot

  39. Deletion Operation Three possible cases to delete a node, x, from a BST 1. The node, x, is a leaf

  40. Deletion Operation 2. The node, x has one child

  41. Delete node pointed to by xSucc as described for cases 1 and 2 K Replace contents of x with inorder successor Deletion Operation • x has two children View remove() function

  42. Question? • Why do we choose inorder predecessor/successor to replace the node to be deleted in case 3?

  43. Implement Delete Operation • void delete(const T& item) //remove the specified item if it exists • Assume you have this function void BST::search2(const T& item, bool& found, BST::BinNodePtr& locptr, BST::BinNodePtr& parent) const { locptr = myRoot; parent = NULL; found = false; while(!found && locptr) { if (item < locptr->data) { parent = locptr; locptr = locptr->left; } else if (item > locptr->data) { parent = locptr; locptr = loc->right; } else found = true; } //end while }

  44. Implement Delete Operation • Search the specified item on the tree. If not found, then return; otherwise go to 2. • Check if the node to be deleted has two child nodes (case 3). If so, find the inorder successor/predecessor, replace the data, and then make the successor/predecessor node the one to be deleted • In order to delete the node, you should keep track of its parent node in step 1 and 2. • Delete the node according to the first two simple cases. Be careful with parent • Release the memory of the node deleted

  45. delete(const T& item) { BinNodePtr parent, x; bool found; search2(item, found, x, parent); //search item on the tree if (!found) return; if (x->left && x->right) { //the case 3 with two child nodes BinNodePtr xsucc =x->right; parent = x; while (xsucc->left) { //find inorder successor parent = xsucc; xsucc = xsucc->left; } x->data = xsucc->data; x = xsucc; } //end if BinNodePtr subtree = x->left; if (!subtree) subtree = x->right; if (!parent) myRoot = subtree; else { if (x == parent->left) parent->left = subtree; else parent->right = subtree; } delete x; }

  46. Questions? • What is T(n) of search algorithm in a BST? Why? • What is T(n) of insert algorithm in a BST? • Other operations?

  47. Problem of Lopsidedness • The order in which items are inserted into a BST determines the shape of the BST • Result in Balanced or Unbalanced trees • Insert O, E, T, C, U, M, P • Insert C, O, M, P, U, T, E

  48. Balanced

  49. Unbalanced

  50. Problem of Lopsidedness • Trees can be totally lopsided • Suppose each node has a right child only • Degenerates into a linked list Processing time affected by "shape" of tree