1. Yan Shi CS/SE 2630 Lecture Notes
10. Binary Search Tree
Yan Shi
CS/SE 2630 Lecture Notes
Partially adopted from C++ Plus Data Structure textbook slides

2. Review: Binary Search in a Sorted Array Based List
Examines the element in the middle of the array. Is it the sought item? If so, stop searching. Is the middle element too small? Then start looking in second half of array. Is the middle element too large? Then begin looking in first half of the array.
Repeat the process in the half of the list that should be examined next.
Stop when item is found, or when there is nowhere else to look and item has not been found.

3. Example
item = 45
info[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]

4. Example
item = 45
info[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
first midPoint last
LESS last = midPoint - 1
GREATER first = midPoint + 1
info[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
first midPoint last

5. item = 45
info[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
first, last
midPoint
GREATER first = midPoint + 1
info[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
first,
midPoint,
last
LESS last = midPoint - 1

6. item = 45
info[0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
last first
first > last found = false

7. Motivations for Binary Search Tree
array-based list:
fast search: binary search O(log2N)
slow insertion/deletion
linear linked list:
fast insertion, deletion
slow search
Can we have fast search, addition and deletion?
Binary search tree

8. What is a tree? Jake’s Pizza Shop Owner Jake Manager Chef Brad Carol
Waitress Waiter Cook Helper
Joyce Chris Max Len

9. A Tree Has a Root Node Owner Jake ROOT NODE Manager Chef Brad Carol
Waitress Waiter Cook Helper
Joyce Chris Max Len

10. Leaf Nodes have No Children
Owner
Jake
Manager Chef Brad Carol
Waitress Waiter Cook Helper
Joyce Chris Max Len

11. A Tree Has Levels Owner LEVEL 0 Jake Manager Chef Brad Carol
Waitress Waiter Cook Helper
Joyce Chris Max Len
LEVEL 0

12. Level One Owner Jake Manager Chef Brad Carol LEVEL 1
Waitress Waiter Cook Helper
Joyce Chris Max Len

13. Level Two Owner Jake Manager Chef Brad Carol LEVEL 2
Waitress Waiter Cook Helper
Joyce Chris Max Len
LEVEL 2

14. A Subtree Owner Jake Manager Chef Brad Carol LEFT SUBTREE OF ROOT NODE
Waitress Waiter Cook Helper
Joyce Chris Max Len
LEFT SUBTREE OF ROOT NODE

15. Binary Tree A binary tree is a structure in which:
Each node can have at most two children, and in which a unique path exists from the root to every other node.
The two children of a node are called the left child and the right child, if they exist.

16. Implementing a Binary Tree with Pointers and Dynamic DataV Q L T E A K S
How many leaf nodes?
descendant of Q?
ancestors of Q?

17. Binary Search Tree (BST)
A special kind of binary tree in which:
Each node contains a distinct data value
The key values in the tree can be compared using “greater than” and “less than”, and
The key value of each node in the tree is less than every key value in its right sub-tree, and greater than every key value in its left sub-tree.
Similar as sorted list, the order is maintained when inserting new items into the tree

18. Example Insert ‘J’ ‘E’ ‘F’ ‘T’ ‘A’ to an empty BST.
Insert ‘A’ ‘E’ ‘F’ ‘J’ ‘T’ to an empty BST.
‘J’
‘E’
‘F’
‘T’
‘A’
The shape of a BST depends on
its key values and
their order of insertion
‘A’
‘E’
‘F’
‘J’
‘T’

19. Exercise ‘J’ ‘E’ ‘T’ ‘A’ ‘H’ ‘M’ ‘K’ ‘P’
Insert nodes containing these values in this order:
‘D’ ‘B’ ‘L’ ‘Q’ ‘S’ ‘V’ ‘Z’

20. How many nodes do we have?
‘J’
‘E’
‘T’
‘A’
‘H’
‘M’
‘V’
‘D’
‘K’
‘Z’
‘P’
‘B’
‘L’
‘Q’
‘S’

21. Count Nodes in a BST CountNodes Algorithm v1:
if left(tree) is NULL AND right(tree) is NULL
return 1
else
return CountNodes(Left(tree)) +
CountNodes(Right(tree)) + 1
What if Left(tree) is NULL?

22. Count Nodes in a BST CountNodes Algorithm v2:
if (Left(tree) is NULL) AND (Right(tree) is NULL)
return 1
else if Left(tree) is NULL
return CountNodes(Right(tree)) + 1
else if Right(tree) is NULL
return CountNodes(Left(tree)) + 1
else
return CountNodes(Left(tree)) +
CountNodes(Right(tree)) + 1
What if the initial tree is NULL?

23. Count Nodes in a BST CountNodes Algorithm v3:
if tree is NULL
return 0
else if (Left(tree) is NULL) AND (Right(tree) is NULL)
return 1
else if Left(tree) is NULL
return CountNodes(Right(tree)) + 1
else if Right(tree) is NULL
return CountNodes(Left(tree)) + 1
else
return CountNodes(Left(tree)) +
CountNodes(Right(tree)) + 1
Can we simplify it somehow?

24. Count Nodes in a BST CountNodes Algorithm Final Version:
if tree is NULL
return 0
else
return CountNodes(Left(tree)) +
CountNodes(Right(tree)) + 1
Does it cover all possible situations?

25. Is ‘F’ in the binary search tree?
‘J’
‘E’
‘T’
‘A’
‘H’
‘M’
‘V’
‘D’
‘K’
‘Z’
‘P’
‘B’
‘L’
‘Q’
‘S’

26. Search an item in a BST Search Algorithm
Similar algorithm for retrieving an item given its key
Given a BST tree and an item to find:
if tree is NULL
found = false
else if ( item < current node’s info )
Search( left(tree), item )
else if ( item > current node’s info )
Search( right(tree), item )
else
found = true

27. Insert an item to a BST Use recursion Base case: General case:
tree is NULL (empty tree or leaf node’s left/right)
CANNOT add  item already in the BST!
General case:
if item < tree->info, Insert item to left(tree)
else if item > tree->info, Insert item to right(tree)

28. The “tree” Parameter
bool BST::Insert( TNode * tree, Type item ) { if ( tree == NULL ) tree = new TNode( item ); return true; } else if ( item < tree->info ) return Insert( tree->left, item ); else if ( item > tree->info ) return Insert( tree->right, item ); else return false; } //Is it correct?

29. The “tree” parameter
Treat “tree” as a pointer pointing within the tree!
What happens when creating a new node?
tree become a real address instead of NULL!
Make “tree” a reference parameter!

30. The “tree” Parameter
bool BST::Insert( TNode * & tree, Type item ) { if ( tree == NULL ) tree = new TNode( item ); return true; } else if ( item < tree->info ) return Insert( tree->left, item ); else if ( item > tree->info ) return Insert( tree->right, item ); else return false; } //tree is a reference parameter!

31. Delete an item in the BST
Use recursion
Base case:
If item's key matches key in Info(tree), delete node pointed to by tree.
If tree is empty, cannot delete.
General case:
if item < tree->info, Delete item in left(tree)
else if item > tree->info, Delete item in right(tree)
How to delete a node?

32. Code for Delete
bool BST::Delete( TNode * & tree, Type item ) { if ( tree == NULL ) return false; if ( item < tree->info ) return Delete( tree->left, item ); else if ( item > tree->info ) return Delete( tree->right, item ); else DeleteNode( tree ); return true; }
we are changing the structure of the tree!
// How to do this?

33. Deleting a Leaf Node

34. Deleting a Node with One Child

35. Deleting a Node with Two Children

36. Delete a Node in BST Algorithm: Indirect Recursion!
if (Left(tree) is NULL) AND (Right(tree) is NULL)
Set tree to NULL
else if Left(tree) is NULL
Set tree to Right(tree)
else if Right(tree) is NULL
Set tree to Left(tree)
else
Find predecessor
Set Info(tree) to Info(predecessor)
Delete predecessor
Indirect Recursion!

37. How to find a predecessor?
Algorithm:
if ( tree->left != NULL ) {
tree = tree->left;
while (tree->right != NULL)
tree = tree->right;
predecessor = tree->info; }

38. Tree Traversal In-order pre-order post-order from smallest to largest
current info first
then left
then right
post-order
left first
then current info

39. In-Order Print Algorithm: How about pre-order and post-order print?
if tree is not NULL
InOrderPrint( left(tree) )
print tree->info
InOrderPrint( right(tree) )

40. Big-3 for BST Destructor: Copy constructor and assignment operator:
need to delete all node
traverse the tree in which order?
Copy constructor and assignment operator:
How to copy a tree?
How to implement operator=?
post-order!
pre-order!

41. Iterative Solutions for BST
Binary Search: Has( item )
Insert: Add( item )
Delete: Remove( item )

42. Iterative Solution for Search
Algorithm for search: Has( item )
Set nodePtr to tree
while more elements to search
if item < Info(nodePtr)
Set nodePtr to Left(nodePtr)
else if item > Info(nodePtr)
Set nodePtr to Right(nodePtr)
else
return true
return false

43. Iterative Solution for Insert
create a node to contain the new item
find the insertion place
similar as the previous search
need to keep track of parent as well
insert new node
insert as either the left or right child of parent

44. Example: Insert
Insert 13 to the tree

45. Insert 13 to the tree

46. Insert 13 to the tree

47. Insert 13 to the tree

48. Insert 13 to the tree

49. Iterative Solution for Delete
TNode * node, * parent; bool found = false; FindNode( item, found, node, parent ); if ( found == false ) return false; if ( node == root ) DeleteNode( root ); else if ( parent->left == node ) DeleteNode( parent->left ); else DeleteNode( parent->right ); return true;
//why parent->left?
is it really iterative if we use the previous DeleteNode function?

50. nodePtr and parentPtr Are External to the Tree

51. parentPtr is External to the Tree, but parentPtr-> left is an Actual Pointer in the Tree

52. Full and Complete BST
Full Binary Tree: A binary tree in which all of the leaves are on the same level and every non-leaf node has two children
Complete Binary Tree: A binary tree that is either full or full through the next-to-last level, with the leaves on the last level as far to the left as possible

53. Examples

54. Array Representation of BST
For any tree node nodes[index]
left child:
nodes[index*2 + 1]
right child:
nodes[index*2 + 2]
parent:
nodes[(index – 1)/2]

55. Array Representation of Non-Complete BST
Use dummy values to fill the space