Chapter 8Binary Search Tree

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Fall 2010

Binary Trees

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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

Trees: level and height

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• Level: distance of a node from the root• Height: the maximum level

Trees

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Why is this not atree?

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

Descendants

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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.

Ancestors

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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.

Binary Trees Come in Different Shapes

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How many structurally different binary trees can be madefrom 2 nodes? 4 nodes? 6 nodes?

Binary Tree Facts• Max. number of nodes at Nth level : 2N.

– 0th level: root node: 1– 1st level: 2– Double as level increases by 1

• Suppose f(n) denotes the maximum number of nodes at the nth level:– f(0)=1– f(n)=2*f(n-1) for n>=1

• A recursive formula! • f(n)=2n

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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:

– 20+21+22+…+2N= 2N+1-1

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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: log2N + 1

• Max. number of levels with N nodes ?– One node at each level => degenerates to a

linked list– The answer is: N

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Binary Search Trees (BST)

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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)

Binary Search Trees

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Each node is the root of a subtree rooted at the node

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)

Tree node in BST

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Can you define a structure to represent a binary tree node ?

Recursive Count

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Let’s start by counting the number of nodes in a tree:

Size?

Base cases(s)?

General case(s)?

Recursive Count: version 1

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if (Left(tree) is NULL) AND (Right(tree) is NULL)

return 1

else

return Count(Left(tree)) + Count(Right(tree)) + 1

Apply tothese trees:

Recursive Count: version 2

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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

Recursive Count: version 3

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if (tree is NULL) return 0if (Left(tree) is NULL) AND (Right(tree) is NULL)

return 1else if (Left(tree) is NULL)

return Count(Right(tree)) + 1else if (Right(tree) is NULL)

return Count(Left(tree)) + 1else return Count(Left(tree)) + Count(Right(tree)) + 1

Recursive Count: version 4

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if (tree is NULL)return 0

elsereturn Count(Left(tree)) + Count(Right(tree)) + 1

Recursive Count: implementation

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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?

Recursive Search

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‘J’

‘E’

‘A’ ‘H’

‘T’

‘M’

‘K’

‘V’

‘P’ ‘Z’‘D’

‘Q’‘L’‘B’

‘S’Are ‘D’, ‘Q’, and ‘N’ in the tree?

Recursive Search

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Retrieve(tree, item, found)

Size?

Base case(s)?

General case(s)?

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

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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

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‘J’

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?

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‘J’

‘E’

Shape of BST• Trace path to insert ‘F’

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‘J’

‘E’

‘F’

Shape of BST

• Trace path to insert ‘T’

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‘J’

‘E’

‘F’

‘T’

Shape of BST

• Trace path to insert ‘A’

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‘J’

‘E’

‘F’

‘T’

‘A’

Shape of BST• Now build tree by inserting ‘A’ ‘E’ ‘F’ ‘J’ ‘T’ in that order

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And the moral is?

Recursive Insertion

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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 locationSet right and left to NULLSet info to item

else if (item is larger than tree->info) Insert(tree->right, item)else Insert(tree->left, item)

Recursive Insertion

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Recursive Insertion

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Insert item 12

Recursive Insertion

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How mustthe tree

bepassed?

Recursive Insertionvoid 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 < tree->info) Insert(tree->left, item); else Insert(tree->right, item); }

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Deleting a Leaf Node

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Deleting a Node with One Child

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Deleting a Node with Two Children

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Delete an existing item from BST

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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.

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, …

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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, …

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Recursive DeletionDeleteItem(tree, item)if (Left(tree) is NULL) AND (Right(tree) is NULL) // delete ‘Z’

Set tree to NULLelse 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 predecessorSet Info(tree) to Info(predecessor)Delete predecessor

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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.

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

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

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Recursive Deletionvoid 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);

}}

45Tracing this function using various examples…

Find Predecessorvoid 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.

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Recursive Deletion

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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

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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))

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A B C D E F G

Preorder Traversal

• Preorder traversal visits the root first.

• PreOrder(tree)

if tree is not NULL1. Visit Info(tree)

2. Preorder(Left(tree))

3. Preorder(Right(tree)) D B A C F E G

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Postorder Traversal

• Postorder traversal visits the root last. • PostOrder(tree)

if tree is not NULL1. Postorder(Left(tree))

2. Postorder(Right(tree))

3. Visit Info(tree)

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A C B E G F D

Traversals

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Printing the Tree

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Traversal Algorithm : Inorder traversal