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Chapter 8 Binary Search Tree 1 Fall 2010. Binary Trees 2 A structure with: i) a unique starting node (the root), in which ii) each node has up to two.

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Presentation on theme: "Chapter 8 Binary Search Tree 1 Fall 2010. Binary Trees 2 A structure with: i) a unique starting node (the root), in which ii) each node has up to two."— Presentation transcript:

1 Chapter 8 Binary Search Tree 1 Fall 2010

2 Binary Trees 2 A structure with: i) a unique starting node (the root), in which ii) each node has up to two child nodes, and iii) a unique path exists from the root to every other node Root Top node of a tree structure; a node with no parent Leaf Node A tree node that has no children

3 Trees: level and height 3 Level: distance of a node from the root Height: the maximum level

4 Trees 4 Why is this not a tree? A tree is a structure with i) a unique starting node (the root), in which ii) each node can have up to two child nodes and iii) a unique path exists from the root to every other node

5 Descendants 5 Q V T K S A E L How many descendants does Q have? Descendant of a node is a child of the node, and any child of the children, etc.

6 Ancestors 6 Q V T K S A E L How many ancestors does S have? Ancestor of a node: a parent of the node, the parent of the parent, etc.

7 Binary Trees Come in Different Shapes 7 How many structurally different binary trees can be made from 2 nodes? 4 nodes? 6 nodes?

8 Binary Tree Facts Max. number of nodes at N th level : 2 N. – 0 th level: root node: 1 – 1st level: 2 – Double as level increases by 1 Suppose f(n) denotes the maximum number of nodes at the n th level: – f(0)=1 – f(n)=2*f(n-1) for n>=1 A recursive formula! f(n)=2 n 8

9 Binary Tree Facts Max. total number of nodes in a tree of height N – All levels are full, so add the max. number of nodes at each level together: – …+2 N = 2 N

10 Binary Tree Facts Given a binary tree with N nodes, what is the min. number of levels? – Try to fill in each level – The answer is: log 2 N + 1 Max. number of levels with N nodes ? – One node at each level => degenerates to a linked list – The answer is: N 10

11 Binary Search Trees (BST) 11 A BST is a binary tree with a search property. A binary tree in which, for each node: key value in the node is greater than the key value in its left child and any of the left child’s descendents (left sub-tree) key value in the node is less than the key value in its right child and any of the right child’s descendents (right sub-tree)

12 Binary Search Trees 12 Each node is the root of a subtree rooted at the node

13 Binary Search Tree ADT Application level: same as List Logic Level – void MakeEmpty() – bool IsEmpty() – bool IsFull() – int GetLength() – RetrieveItem(ItemType &item, bool &found) – InsertItem (ItemType item) – DeleteItem (ItemType item) – Print(ofstream &outFile) – ResetTree(OrderType order) – GetNextItem (ItemType &item, OrderType order,bool &finished)

14 Tree node in BST 14 Can you define a structure to represent a binary tree node ?

15 Recursive Count 15 Let’s start by counting the number of nodes in a tree: Size? Base cases(s)? General case(s)?

16 Recursive Count: version 1 16 if (Left(tree) is NULL) AND (Right(tree) is NULL) return 1 else return Count(Left(tree)) + Count(Right(tree)) + 1 Apply to these trees:

17 Recursive Count: version 2 17 if (Left(tree) is NULL) AND (Right(tree) is NULL) return 1 else if (Left(tree) is NULL) return Count(Right(tree)) + 1 else if (Right(tree) is NULL) return Count(Left(tree)) + 1 else return Count(Left(tree)) + Count(Right(tree)) + 1 Apply to an empty tree

18 Recursive Count: version 3 18 if (tree is NULL) return 0 if (Left(tree) is NULL) AND (Right(tree) is NULL) return 1 else if (Left(tree) is NULL) return Count(Right(tree)) + 1 else if (Right(tree) is NULL) return Count(Left(tree)) + 1 else return Count(Left(tree)) + Count(Right(tree)) + 1

19 Recursive Count: version 4 19 if (tree is NULL) return 0 else return Count(Left(tree)) + Count(Right(tree)) + 1

20 Recursive Count: implementation 20 int TreeType::GetLength() const { return Count(root); ) int TreeType::Count(TreeNode* tree) const { if (tree == NULL) return 0; else return Count(tree->left) + Count(tree->right) + 1; } Why do we need two functions?

21 Recursive Search 21 ‘J’ ‘E’ ‘A’ ‘H’ ‘T’ ‘M’ ‘K’ ‘V’ ‘P’ ‘Z’‘D’‘Q’‘L’‘B’‘S’ Are ‘D’, ‘Q’, and ‘N’ in the tree?

22 Recursive Search 22 Retrieve(tree, item, found) Size? Base case(s)? General case(s)?

23 Recursive Search void TreeType::Retrieve(TreeNode* tree, ItemType& item, bool& found) const { if (tree == NULL) found = false; //base case else if (item info) Retrieve(tree->left, item, found); else if (item > tree->info) Retrieve(tree->right, item, found); else //base case { item = tree->info; found = true; } 23

24 Shape of BST Shape depends on the order of item insertion Insert the elements ‘J’ ‘E’ ‘F’ ‘T’ ‘A’ in that order The first value inserted is always put in the root 24 ‘J’

25 Shape of BST Thereafter, each value to be inserted: – compares itself to the value in the root node – moves left it is less or – moves right if it is greater When does the process stop? 25 ‘J’ ‘E’

26 Shape of BST Trace path to insert ‘F’ 26 ‘J’ ‘E’ ‘F’

27 Shape of BST Trace path to insert ‘T’ 27 ‘J’ ‘E’ ‘F’ ‘T’

28 Shape of BST Trace path to insert ‘A’ 28 ‘J’ ‘E’ ‘F’ ‘T’ ‘A’

29 Shape of BST Now build tree by inserting ‘A’ ‘E’ ‘F’ ‘J’ ‘T’ in that order 29 And the moral is?

30 Recursive Insertion 30 Insert an item into a tree Where does each new node get inserted? Insert(tree, item) if (tree is NULL) Get a new node and insert at this location Set right and left to NULL Set info to item else if (item is larger than tree->info) Insert(tree->right, item) else Insert(tree->left, item)

31 Recursive Insertion 31

32 Recursive Insertion 32 Insert item 12

33 Recursive Insertion 33 How must the tree be passed?

34 Recursive Insertion void TreeType::Insert(TreeNode* & tree, ItemType item) { if (tree == NULL) { // Insertion place found. tree = new TreeNode; tree->right = NULL; tree->left = NULL; tree->info = item; } else if (item info) Insert(tree->left, item); else Insert(tree->right, item); } 34

35 Deleting a Leaf Node 35

36 Deleting a Node with One Child 36

37 Deleting a Node with Two Children 37

38 Delete an existing item from BST 38 Can you summarize the three deletion cases? 1.Deleting a leaf node. 2.Deleting a node with only one child. 3.Deleting a node with two children.

39 Predecessor Predecessor: the element whose key immediately precedes (less than) the key of item – If the item node has a left child, the largest element in the left subtree (right-most child) – If the item has no left child, … 39

40 Successor Successor: the element whose key immediately follows (greater than) the key of item – If the item node has two children, the smallest element in the right subtree (left-most child) – If the item node has no right child, … 40

41 Recursive Deletion DeleteItem(tree, item ) if (Left(tree) is NULL) AND (Right(tree) is NULL) // delete ‘Z’ Set tree to NULL else if (Left(tree) is NULL AND (Right(tree)) is not NULL) delete ‘R’ Set tree to Right(tree) else if (Right(tree) is NULL AND (Left(tree)) is not NULL) Set tree to Left(tree) else // delete ‘Q’, maintain a binary search tree Find predecessor Set Info(tree) to Info(predecessor) Delete predecessor 41

42 Recursive Deletion TreeType::DeleteItem (ItemType item) – deletes item from the current object (a tree object) void Delete( TreeNode*& tree, ItemType item) – deletes item from a tree rooted at tree void DeleteNode( TreeNode*& tree) – deletes node pointed to by tree from a BST void GetPredecessor( TreeNode* tree, ItemType& data) – finds data’s predecessor – the largest item in data’s left subtree and saves the info into data. 42

43 Recursive Deletion void TreeType::DeleteItem(ItemType item) { Delete(root, item); } //Calls the recursive function Delete to //delete item from tree. 43

44 Recursive Deletion // first, find which node should be deleted. void TreeType::Delete(TreeNode*& tree, ItemType item) { if (item info) Delete(tree->left, item); else if (item > tree->info) Delete(tree->right, item); else DeleteNode(tree); // Node found } 44

45 Recursive Deletion void TreeType::DeleteNode(TreeNode*& tree) { ItemType data; TreeNode* tempPtr; tempPtr = tree; if ( tree->left == NULL) { tree = tree->right; delete tempPtr; } else if (tree->right == NULL){ tree = tree->left; delete tempPtr; }else{ GetPredecessor(tree->left, data); tree->info = data; Delete(tree->left, data); } 45 Tracing this function using various examples…

46 Find Predecessor void TreeType::GetPredecessor( TreeNode* tree, ItemType& data) { //the largest item is located in its rightmost node. while (tree->right != NULL) tree = tree->right; data = tree->info; } This function could be named GetLargestItem() as it returns the largest item stored in tree rooted at tree. 46

47 Recursive Deletion 47

48 Traversals Tree Traversal : visiting all the nodes of a tree – Depth-First Traversal, Breadth-First Traversal Depth-First Traversal – Inorder Traversal – Preorder Traversal – Postorder Traversal 48

49 Inorder Traversal Inorder traversal visits the root in between visiting the left and right subtrees: – Inorder traversal only makes sense for binary trees. Inorder(tree) if tree is not NULL 1.Inorder(Left(tree)) 2.Visit Info(tree) 3.Inorder(Right(tree)) 49 A B C D E F G

50 Preorder Traversal Preorder traversal visits the root first. PreOrder(tree) if tree is not NULL 1.Visit Info(tree) 2.Preorder(Left(tree)) 3.Preorder(Right(tree)) D B A C F E G 50

51 Postorder Traversal Postorder traversal visits the root last. PostOrder(tree) if tree is not NULL 1.Postorder(Left(tree)) 2.Postorder(Right(tree)) 3.Visit Info(tree) 51 A C B E G F D

52 Traversals 52

53 Printing the Tree 53 Traversal Algorithm : Inorder traversal


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