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Binary Search Trees (BST)

Binary Search Trees (BST). Binary Search Trees (BST). Binary tree (but not BST). Binary Search Tree (Balanced). Binary Search Tree (Un-balanced). Definitions. Binary tree Each node has at most two children Binary Search Tree (BST): Is a binary tree where:

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Binary Search Trees (BST)

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  1. Binary Search Trees(BST)

  2. Binary Search Trees (BST) Binary tree (but not BST) Binary Search Tree (Balanced) Binary Search Tree (Un-balanced)

  3. Definitions • Binary tree • Each node has at most two children • Binary Search Tree (BST): Is a binary tree where: • Left subtree is always less than the node • Right subtree is always greater than or equal the node

  4. Height of BST • Can be balanced • Left of the root ≈ Right of the root • In this case the height O(Log n) • Can be un-balanced • In this case the height is worst case O(n)

  5. BST Traversal Three main types • Pre-Order Traversal • Post-Order Traversal • In-Order Traversal

  6. Traversing a Binary Search Tree (BST)Pre-Order

  7. Outline of Pre-Order Traversal • For each node • Do the work first (Current) • Traverse Left • Traverse Right • Work can be anything (E.g., print)

  8. Traverse the tree “Pre-order”: • Visit the current and do work • Visit the tree’s left sub-tree • Visit right sub-tree

  9. Pre-Order Traversal Procedure procedure Pre_Order(Ptr Pointer to Tree_Node) • if( Ptr <> NIL ) then • Do_Whatever( Ptr->data ) • Pre_Order( Ptr->left_child ) • Pre_Order( Ptr->right_child ) • endif • endprocedure Two recursive calls

  10. 7 97 94 44 14 1 36 67 9 22 3 Proc PreOrderPrint(pointer) pointer NOT NIL? print(data) PreOrderPrint(left child) PreOrderPrint(right child) root P L R Output 22, 9, 3, 1, 7, 14, 67, 36, 44, 94, 97

  11. Traversing a Binary Search Tree (BST)Post-Order

  12. Outline of Post-Order Traversal • For each node • Traverse Left • Traverse Right • Do the work (Current)

  13. Post-Order Traversal Procedure procedure Post_Order(Ptr Pointer to Tree_Node) • if( Ptr <> NIL ) then • Post_Order( Ptr->left_child ) • Post_Order( Ptr->right_child ) • Do_Whatever( Ptr->data ) • endif • endprocedure Two recursive calls

  14. 7 97 94 44 14 1 36 67 9 22 3 Proc PostOrderPrint(pointer) pointer NOT NIL? PostOrderPrint(left child) PostOrderPrint(right child) print(data) root L R P Output 1, 7, 3, 14, 9, 44, 36, 97, 94, 67, 22

  15. Traversing a Binary Search Tree (BST)In-Order

  16. Outline of In-Order Traversal • For each node • Traverse Left • Do the work (Current) • Traverse Right

  17. In-Order Traversal Procedure procedure In_Order(Ptr Pointer to Tree_Node) • if( Ptr <> NIL ) then • In_Order( Ptr->left_child ) • Do_Whatever( Ptr->data ) • In_Order( Ptr->right_child ) • endif • endprocedure Two recursive calls

  18. 94 97 7 44 14 1 36 67 9 22 3 ProcInOrderPrint(pointer) pointer NOT NIL? InOrderPrint(left child) print(data) InOrderPrint(right child) root L P R Output 1, 3, 7, 9, 14, 22, 36, 44, 67, 94, 97 In-order traversal produces a sorted list

  19. 10 33 20 60 Notes • The three traversal types (Pre-, In-, Post-)Order apply to all types of binary trees • Whether binary search tree or not • In-Order traversal is applicable only for binary trees • Not for N-way trees, where N > 2 • 3-way tree • Pre-Order and Post-Order traversal are applicable • In-Order traversal is not applicable

  20. Analysis of Tree Traversal • Assume the tree has n nodes • Any the three types of traversal is linearO(n) • Each node is visited only once (E.g., printing is done once)

  21. Searching

  22. Problem Definition • Input • Binary tree • Value k to search for Algorithm depdens on whether it is binary search tree (BST) or not

  23. Searching Binary Trees (Not BST) Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure If the current node contains k, return true Call the left and right subtrees recursively (Will check the L.H.S first)

  24. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure

  25. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure

  26. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure

  27. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure

  28. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure

  29. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure

  30. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure

  31. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure Similar to Pre-Order traversal - Check the node first - Check the L.H.S - Check the R.H.S

  32. Time Complexity For Binary Tree Searching Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else returnTree-Search(ptr->left, k) || Tree-Search(ptr->right, k) end if; End procedure • How many times each node is checked? • At most once • Searching a binary tree is linear  O(n)

  33. Find K in Binary Search Trees (BST) • Should make use of the BST property to search in an efficient way Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else if (k < ptr.data) returnTree-Search(ptr->left, k) else return Tree-Search(ptr->right, k) end if; End procedure

  34. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else if (k < ptr.data) returnTree-Search(ptr->left, k) else return Tree-Search(ptr->right, k) end if; End procedure

  35. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else if (k < ptr.data) returnTree-Search(ptr->left, k) else return Tree-Search(ptr->right, k) end if; End procedure

  36. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else if (k < ptr.data) returnTree-Search(ptr->left, k) else return Tree-Search(ptr->right, k) end if; End procedure

  37. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else if (k < ptr.data) returnTree-Search(ptr->left, k) else return Tree-Search(ptr->right, k) end if; End procedure

  38. Example: Search For 6 Tree-Search(ptr pointer to current root, value k) if (ptr == NULL) return false; else if (ptr.data = k) return true; else if (k < ptr.data) returnTree-Search(ptr->left, k) else return Tree-Search(ptr->right, k) end if; End procedure Not Found

  39. Time Complexity For BST Searching • In BST search we follow ONLY on path from root to a leaf • What is the Worst Case Complexity? • O(h), where h is the tree height • If BST is balanced  O (Log n) • If BST is not balanced  O (n)

  40. Insertions & Deletions in BST

  41. Insertion of V in BST • Search for the correctempty place to put the new node • Correct: if V >= current node  move right if V < current node  move left • Empty:Find link with Null value

  42. Example: Insert value 6 • Search for the correctempty place to put the new node • Correct: if V >= current node  move right if V < current node  move left • Empty:Find link with Null value

  43. Example: Insert value 6 • Search for the correctempty place to put the new node • Correct: if V >= current node  move right if V < current node  move left • Empty:Find link with Null value

  44. Example: Insert value 6 • Search for the correctempty place to put the new node • Correct: if V >= current node  move right if V < current node  move left • Empty:Find link with Null value

  45. Example: Insert value 6 • Search for the correctempty place to put the new node • Correct: if V >= current node  move right if V < current node  move left • Empty:Find link with Null value

  46. Example: Insert value 6 • Search for the correctempty place to put the new node • Correct: if V >= current node  move right if V < current node  move left • Empty:Find link with Null value

  47. Example: Insert value 6 • Search for the correctempty place to put the new node • Correct: if V >= current node  move right if V < current node  move left • Empty:Find link with Null value Time complexity for insertion is O(h), h: is the tree height

  48. Deletion of V from BST • Search for V in the tree • Not Found  We are done • Found • Case 1: Node V has no children • Delete V • Case 2: Node V has one child • The child takes the place of V • Case 3: Node V has two children • Find the largest node smaller than V  Say X • Put X in the place of V How to find this value (Successor of V)??

  49. Find the Successor Of Given Node V • Successor of V: is the smallest value larger than V • How to find it • Go to the R.H.S child node • Follow the left link all the way (until Null) • Successor of 22 • 36 • Successor of 3 • 6 Time complexity for finding the successor is O(h), h is the tree height

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