BinarySearchTrees.ppt

Binary Search Trees
CS 302 – Data Structures
Chapter 8

What is a binary tree?
• Property 1: each node can have up to two
successor nodes.

• Property 2: a unique path exists from the
root to every other node
What is a binary tree? (cont.)
Not a valid binary tree!

Some terminology
• The successor nodes of a node are called its children
• The predecessor node of a node is called its parent
• The "beginning" node is called the root (has no parent)
• A node without children is called a leaf

Some terminology (cont’d)
• Nodes are organize in levels (indexed from 0).
• Level (or depth) of a node: number of edges in the path
from the root to that node.
• Height of a tree h: #levels = L
(Warning: some books define h
as #levels-1).
• Full tree: every node has exactly
two children and all the
leaves are on the same level.
not full!

What is the max #nodes
at some level l?
The max #nodes at level is
l 2l
0
2

1
2

2
2

3
2

where l=0,1,2, ...,L-1

What is the total #nodes N
of a full tree with height h?
0 1 1
1
0 1 1 1
1
0
2 2 ... 2 2 1
...
n
h h
n
n i x
x
i
N
x x x x


 


     
    

using the geometric series:
l=0 l=1 l=h-1

What is the height h
of a full tree with N nodes?
2 1
2 1
log( 1) (log )
h
h
N
N
h N O N
 
  
   

Why is h important?
• Tree operations (e.g., insert, delete, retrieve
etc.) are typically expressed in terms of h.
• So, h determines running time!

•What is the max height of a tree with
N nodes?
•What is the min height of a tree with
N nodes?
N (same as a linked list)
log(N+1)

How to search a binary tree?
(1) Start at the root
(2) Search the tree level
by level, until you find
the element you are
searching for or you reach
a leaf.
Is this better than searching a linked list?
No  O(N)

Binary Search Trees (BSTs)
• Binary Search Tree Property:
The value stored at
a node is greater than
the value stored at its
left child and less than
the value stored at its
right child

In a BST, the value
stored at the root of
a subtree is greater
than any value in its
left subtree and less
than any value in its
right subtree!

Where is the
smallest element?
Ans: leftmost element
Where is the largest
element?
Ans: rightmost element

How to search a binary search tree?
(1) Start at the root
(2) Compare the value of
the item you are
searching for with
the value stored at
the root
(3) If the values are
equal, then item
found; otherwise, if it
is a leaf node, then
not found

(4) If it is less than the value
stored at the root, then
search the left subtree
(5) If it is greater than the
value stored at the root,
then search the right
subtree
(6) Repeat steps 2-6 for the
root of the subtree chosen
in the previous step 4 or 5

Is this better than searching
a linked list?
Yes !! ---> O(logN)

Tree node structure
template<class ItemType>
struct TreeNode<ItemType> {
ItemType info;
TreeNode<ItemType>* left;
TreeNode<ItemType>* right;
};

Binary Search Tree Specification
#include <fstream.h>
struct TreeNode<ItemType>;
enum OrderType {PRE_ORDER, IN_ORDER, POST_ORDER};
class TreeType {
public:
TreeType();
~TreeType();
TreeType(const TreeType<ItemType>&);
void operator=(const TreeType<ItemType>&);
void MakeEmpty();
bool IsEmpty() const;
bool IsFull() const;
int NumberOfNodes() const;

void RetrieveItem(ItemType&, bool& found);
void InsertItem(ItemType);
void DeleteItem(ItemType);
void ResetTree(OrderType);
void GetNextItem(ItemType&, OrderType, bool&);
void PrintTree(ofstream&) const;
private:
TreeNode<ItemType>* root;
};
Binary Search Tree Specification
(cont.)

Function NumberOfNodes
• Recursive implementation
#nodes in a tree =
#nodes in left subtree + #nodes in right subtree + 1
• What is the size factor?
Number of nodes in the tree we are examining
• What is the base case?
The tree is empty
• What is the general case?
CountNodes(Left(tree)) + CountNodes(Right(tree)) + 1

int TreeType<ItemType>::NumberOfNodes() const
{
return CountNodes(root);
}
int CountNodes(TreeNode<ItemType>* tree)
{
if (tree == NULL)
return 0;
else
return CountNodes(tree->left) + CountNodes(tree->right) + 1;
}
Function NumberOfNodes (cont.)
O(N)
Running Time?

Function RetrieveItem
• What is the size of the problem?
• What is the base case(s)?
1) When the key is found
2) The tree is empty (key was not found)
Search in the left or right subtrees

template <class ItemType>
void TreeType<ItemType>:: RetrieveItem(ItemType& item, bool& found)
{
Retrieve(root, item, found);
}
void Retrieve(TreeNode<ItemType>* tree, ItemType& item, bool& found)
{
if (tree == NULL) // base case 2
found = false;
else if(item < tree->info)
Retrieve(tree->left, item, found);
else if(item > tree->info)
Retrieve(tree->right, item, found);
else { // base case 1
item = tree->info;
found = true;
}
}
Function RetrieveItem (cont.)
O(h)
Running Time?

Function
InsertItem
• Use the
binary search
tree property
to insert the
new item at
the correct
place

Function
InsertItem
(cont.)
• Implementing
insertion
resursively
e.g., insert 11

The tree is empty
Choose the left or right subtree
Function InsertItem (cont.)

void TreeType<ItemType>::InsertItem(ItemType item)
{
Insert(root, item);
}
void Insert(TreeNode<ItemType>*& tree, ItemType item)
{
if(tree == NULL) { // base case
tree = new TreeNode<ItemType>;
tree->right = NULL;
tree->left = NULL;
tree->info = item;
}
else if(item < tree->info)
Insert(tree->left, item);
else
Insert(tree->right, item);
}
O(h)
Running Time?

Insert 11

• Yes, certain orders might produce very
unbalanced trees!
Does the order of inserting
elements into a tree matter?

Does the
order of
inserting
elements
into a tree
matter?
(cont.)

• Unbalanced trees are not desirable because
search time increases!
• Advanced tree structures, such as red-black
trees, guarantee balanced trees.
Does the order of inserting elements
into a tree matter? (cont’d)

Function DeleteItem
• First, find the item; then, delete it
• Binary search tree property must be
preserved!!
• We need to consider three different cases:
(1) Deleting a leaf
(2) Deleting a node with only one child
(3) Deleting a node with two children

(2) Deleting a node with
only one child

(3) Deleting a node with two
children

• Find predecessor (i.e., rightmost node in the
left subtree)
• Replace the data of the node to be deleted
with predecessor's data
• Delete predecessor node
(3) Deleting a node with two
children (cont.)

Key to be deleted was found
Choose the left or right subtree
Function DeleteItem (cont.)

void TreeType<ItmeType>::DeleteItem(ItemType item)
{
Delete(root, item);
}
void Delete(TreeNode<ItemType>*& tree, ItemType item)
{
if(item < tree->info)
Delete(tree->left, item);
Delete(tree->right, item);
else
DeleteNode(tree);
}

template <class ItemType>
void DeleteNode(TreeNode<ItemType>*& tree)
{
ItemType item;
TreeNode<ItemType>* tempPtr;
tempPtr = tree;
if(tree->left == NULL) { // right child
tree = tree->right;
delete tempPtr;
}
else if(tree->right == NULL) { // left child
tree = tree->left;
delete tempPtr;
}
else {
GetPredecessor(tree->left, item);
tree->info = item;
}
}
2 children
0 children or
1 child
0 children or
1 child

void GetPredecessor(TreeNode<ItemType>* tree, ItemType& item)
{
while(tree->right != NULL)
tree = tree->right;
item = tree->info;
}

void TreeType<ItmeType>::DeleteItem(ItemType item)
{
Delete(root, item);
}
void Delete(TreeNode<ItemType>*& tree, ItemType item)
{
if(item < tree->info)
Delete(tree->right, item);
else
DeleteNode(tree);
}
Running Time?
O(h)

Tree Traversals
There are mainly three ways to traverse a tree:
1) Inorder Traversal
2) Postorder Traversal
3) Preorder Traversal

46
Inorder Traversal: A E H J M T Y
‘J’
‘E’
‘A’ ‘H’
‘T’
‘M’ ‘Y’
tree
Visit left subtree first Visit right subtree last
Visit second

Inorder Traversal
• Visit the nodes in the left subtree, then visit
the root of the tree, then visit the nodes in
the right subtree
Inorder(tree)
If tree is not NULL
Inorder(Left(tree))
Visit Info(tree)
Inorder(Right(tree))
Warning: "visit" implies do something with the value at
the node (e.g., print, save, update etc.).

48
Preorder Traversal: J E A H T M Y
‘J’
‘E’
‘A’ ‘H’
‘T’
‘M’ ‘Y’
tree
Visit left subtree second Visit right subtree last
Visit first

Preorder Traversal
• Visit the root of the tree first, then visit the
nodes in the left subtree, then visit the nodes
in the right subtree
Preorder(tree)
If tree is not NULL
Visit Info(tree)
Preorder(Left(tree))
Preorder(Right(tree))

50
‘J’
‘E’
‘A’ ‘H’
‘T’
‘M’ ‘Y’
tree
Visit left subtree first Visit right subtree second
Visit last
Postorder Traversal: A H E M Y T J

Postorder Traversal
• Visit the nodes in the left subtree first, then
visit the nodes in the right subtree, then visit
the root of the tree
Postorder(tree)
If tree is not NULL
Postorder(Left(tree))
Postorder(Right(tree))
Visit Info(tree)

Tree
Traversals:
another
example

Function PrintTree
• We use "inorder" to print out the node values.
• Keys will be printed out in sorted order.
• Hint: binary search could be used for sorting!
A D J M Q R T

Function PrintTree (cont.)
void TreeType::PrintTree(ofstream& outFile)
{
Print(root, outFile);
}
void Print(TreeNode<ItemType>* tree, ofstream& outFile)
{
if(tree != NULL) {
Print(tree->left, outFile);
outFile << tree->info; // “visit”
Print(tree->right, outFile);
}
}
to overload “<<“ or “>>”, see:
http://www.fredosaurus.com/notes-cpp/oop-friends/overload-io.html

Class Constructor
TreeType<ItemType>::TreeType()
{
root = NULL;
}

Class Destructor
How should we
delete the nodes
of a tree?
Use postorder!
Delete the tree in a
"bottom-up"
fashion

Class Destructor (cont’d)
TreeType::~TreeType()
{
Destroy(root);
}
void Destroy(TreeNode<ItemType>*& tree)
{
if(tree != NULL) {
Destroy(tree->left);
Destroy(tree->right);
delete tree; // “visit”
}
}
postorder

Copy Constructor
How should we
create a copy of
a tree?
Use preorder!

Copy Constructor (cont’d)
TreeType<ItemType>::TreeType(const TreeType<ItemType>&
originalTree)
{
CopyTree(root, originalTree.root);
}
template<class ItemType)
void CopyTree(TreeNode<ItemType>*& copy,
TreeNode<ItemType>* originalTree)
{
if(originalTree == NULL)
copy = NULL;
else {
copy = new TreeNode<ItemType>;
copy->info = originalTree->info;
CopyTree(copy->left, originalTree->left);
CopyTree(copy->right, originalTree->right);
}
}
preorder
// “visit”

ResetTree and GetNextItem
• User needs to specify the tree
traversal order.
• For efficiency, ResetTree stores in
a queue the results of the
specified tree traversal.
• Then, GetNextItem, dequeues the
node values from the queue.
void ResetTree(OrderType);
void GetNextItem(ItemType&,
OrderType, bool&);

Revise Tree Class Specification
enum OrderType {PRE_ORDER, IN_ORDER, POST_ORDER};
class TreeType {
public:
// previous member functions
void PreOrder(TreeNode<ItemType>, QueType<ItemType>&)
void InOrder(TreeNode<ItemType>, QueType<ItemType>&)
void PostOrder(TreeNode<ItemType>, QueType<ItemType>&)
private:
TreeNode<ItemType>* root;
QueType<ItemType> preQue;
QueType<ItemType> inQue;
QueType<ItemType> postQue;
};
new private data
new member
functions

void PreOrder(TreeNode<ItemType>tree,
QueType<ItemType>& preQue)
{
if(tree != NULL) {
preQue.Enqueue(tree->info); // “visit”
PreOrder(tree->left, preQue);
PreOrder(tree->right, preQue);
}
}
ResetTree and GetNextItem (cont.)

void InOrder(TreeNode<ItemType>tree,
QueType<ItemType>& inQue)
{
if(tree != NULL) {
InOrder(tree->left, inQue);
inQue.Enqueue(tree->info); // “visit”
InOrder(tree->right, inQue);
}
}

void PostOrder(TreeNode<ItemType>tree,
QueType<ItemType>& postQue)
{
if(tree != NULL) {
PostOrder(tree->left, postQue);
PostOrder(tree->right, postQue);
postQue.Enqueue(tree->info); // “visit”
}
}

ResetTree
void TreeType<ItemType>::ResetTree(OrderType order)
{
switch(order) {
case PRE_ORDER: PreOrder(root, preQue);
break;
case IN_ORDER: InOrder(root, inQue);
break;
case POST_ORDER: PostOrder(root, postQue);
break;
}
}

GetNextItem
void TreeType<ItemType>::GetNextItem(ItemType& item,
OrderType order, bool& finished)
{
finished = false;
switch(order) {
case PRE_ORDER: preQue.Dequeue(item);
if(preQue.IsEmpty())
finished = true;
break;
case IN_ORDER: inQue.Dequeue(item);
if(inQue.IsEmpty())
finished = true;
break;
case POST_ORDER: postQue.Dequeue(item);
if(postQue.IsEmpty())
finished = true;
break;
}
}

Iterative Insertion and Deletion
• Reading Assignment (see textbook)

Comparing Binary Search Trees to Linear Lists
Big-O Comparison
Operation
Binary
Search Tree
Array-
based List
Linked
List
Constructor O(1) O(1) O(1)
Destructor O(N) O(1) O(N)
IsFull O(1) O(1) O(1)
IsEmpty O(1) O(1) O(1)
RetrieveItem O(logN)* O(logN) O(N)
InsertItem O(logN)* O(N) O(N)
DeleteItem O(logN)* O(N) O(N)
*assuming h=O(logN)

BinarySearchTrees.ppt

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