Binary Search Trees

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Binary Search Trees

• CISC2200, Fall 09

2
Trees
Owner Jake
Manager Chef
Brad Carol Waitress
Waiter Cook
Helper Joyce
Chris
Max Len
Jakes Pizza Shop
3
Trees
Owner Jake
Manager Chef
Brad Carol Waitress
Waiter Cook
Helper Joyce
Chris
Max Len
4
Trees
Owner Jake
Manager Chef
Brad Carol Waitress
Waiter Cook
Helper Joyce
Chris
Max Len
5
Trees
Owner Jake
Manager Chef
Brad Carol Waitress
Waiter Cook
Helper Joyce
Chris
Max Len
LEVEL 0
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Trees
Owner Jake
Manager Chef
Brad Carol Waitress
Waiter Cook
Helper Joyce
Chris
Max Len
7
Trees
Owner Jake
Manager Chef
Brad Carol Waitress
Waiter Cook
Helper Joyce
Chris
Max Len
LEVEL 2
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Trees
Owner Jake
Manager Chef
Brad Carol Waitress
Waiter Cook
Helper Joyce
Chris
Max Len
LEFT SUBTREE OF ROOT NODE
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Trees
Owner Jake
Manager Chef
Brad Carol Waitress
Waiter Cook
Helper Joyce
Chris
Max Len
RIGHT SUBTREE OF ROOT NODE
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Binary Trees
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
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Trees level and height

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

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Trees
Why is this not a tree?
A tree is a structure with i) a unique starting
node (the root), in which ii) each node is
capable of having two child nodes and iii) a
unique path exists from the root to every other
node
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Descendants
Descendant of a node is a child of the node, and
any child of the children, etc.
V
Q
L
T
A
E
K
S
How many descendants does Q have?
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Ancestors
Ancestor of a node a parent of the node, the
parent of the parent, etc.
V
Q
L
T
A
E
K
S
How many ancestors does S have?
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Many different shapes of Binary Trees
How many different shape binary trees can be
made from 2 nodes? 4 nodes? 6 nodes?
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Facts of Binary Tree

• Max. number of nodes at Nth level 2N.
• 0th level root node 1
• 1th level 2
• Double as level increases by 1
• Suppose f(n) denote the maximum number of nodes
  at nth level
• f(0)1
• f(n)2f(n-1) for ngt1
• A recursive formula!
• f(n)2n

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Facts of Binary Tree

• 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
• 2021222N 2N1-1

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Facts of Binary Tree

• Given a binary tree with N node, 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 gt degenerate to a linked
  list
• The answer is N

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Binary Search Trees (BST)

• A BST is a binary tree with 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
  childs descendents (left sub-tree)
• key value in the node is less than the key value
  in its right child and any of the right childs
  descendents (right sub-tree)

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

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Tree node in BST
Can you define a structure to represent a binary
tree node ?
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Recursive Count
Lets start by counting the number of nodes in a
tree Size? Base cases(s)? General case(s)?
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Recursive Count version 1
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
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Recursive Count version 2
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
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Recursive Count version 3
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
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Recursive Count version 4
if (tree is NULL) return 0 else return
Count(Left(tree)) Count(Right(tree)) 1
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Recursive Count implementation
int TreeTypeGetLength() const return
Count(root) ) int TreeTypeCount(TreeNode
tree) const if (tree NULL) return 0
else return Count(tree-gtleft)
Count(tree-gtright) 1
Why do we need two functions?
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Recursive Search

J
T
E
A
V
M
H
P
Are D, Q, and N in the tree?
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Recursive Search
Retrieve(tree, item, found) Size? Base
case(s)? General case(s)?
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Recursive Search

• void TreeTypeRetrieve(TreeNode tree,
• ItemType item, bool found) const
• if (tree NULL)
• found false //base case
• else if (item.ComparedTo(tree-gtinfo) LESS)
  
• Retrieve(tree-gtleft, item, found)
• else if (item.ComparedTo(tree-gtinfo) LARGER)
• Retrieve(tree-gtright, item, found)
• else //base case
• item tree-gtinfo
• 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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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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Shape of BST

• Trace path to insert F

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Shape of BST

• Trace path to insert T

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Shape of BST

• Trace path to insert A

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Shape of BST

• Now build tree by inserting A E F
  J T in that order

And the moral is?
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Recursive Insertion
Insert an item into a tree Where does each new
node get inserted?
Insert(tree, item) if (tree is NULL) Get a new
node Set right and left to NULL Set info to
item Else if item is larger than tree.info
Insert(tree-gtright, item) Else
Insert(tree-gtleft, item)
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Recursive Insertion
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Recursive Insertion
Insert item 12
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Recursive Insertion
How must the tree be passed?
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Recursive Insertion

• void TreeTypeInsert(TreeNode tree, ItemType
  item)
• if (tree NULL)
• // Insertion place found.
• tree new TreeNode
• tree-gtright NULL
• tree-gtleft NULL
• tree-gtinfo item
• else if (item.ComparedTo(tree-gtinfo) LESS)
• Insert(tree-gtleft, item)
• else
• Insert(tree-gtright, item)

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Recursive Deletion
Delete Z
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Recursive Deletion
Delete R
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Recursive Deletion
Delete Q
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Delete an existing item from BST

• Can you summarize the three deletion cases?
• Deleting a leaf node.
• Deleting a node with only one child.
• Deleting a node with two children.

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Predecessor

• Predecessor the element whose key immediately
  precedes (less than) the key of item
• If the item node has 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 child,

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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

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Recursive Deletion

• TreeTypeDeleteItem (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 datas predecessor the largest item in
  datas left subtree and save the info into data.

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Recursive Deletion

• // first, find which node should be deleted.
• void TreeTypeDelete(TreeNode tree,
• ItemType item)
• if (item lt tree-gtinfo)
• Delete(tree-gtleft, item)
• else if (item gt tree-gtinfo)
• Delete(tree-gtright, item)
• else
• DeleteNode(tree) // Node found

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Recursive Deletion

• void TreeTypeDeleteNode(TreeNode tree)
• ItemType data
• TreeNode tempPtr
• tempPtr tree
• if ( tree-gtleft NULL)
• tree tree-gtright
• delete tempPtr
• else if (tree-gtright NULL)
• tree tree-gtleft
• delete tempPtr
• else
• GetPredecessor(tree-gtleft, data)
• tree-gtinfo data
• Delete(tree-gtleft, data)
  

Tracing this function using various examples
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Find Predecessor

• void TreeTypeGetPredecessor( TreeNode tree,
• ItemType data)
• //the largest item is located in its rightmost
  node.
• while (tree-gtright ! NULL)
• tree tree-gtright
• data tree-gtinfo
• This function should be named GetLargestItem() as
  it returns the largest item stored in tree rooted
  at tree
• Why is the code not recursive?
• Can you write the recursive solution?

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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
• Inorder(Left(tree))
• Visit Info(tree)
• Inorder(Right(tree))

A B C D E F G
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Preorder Traversal

• Preorder traversal visits the root first.
• PreOrder(tree)
• if tree is not NULL
• Visit Info(tree)
• Preorder(Left(tree))
• 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 NULL
• Postorder(Left(tree))
• Postorder(Right(tree))
• Visit Info(tree)

A C B E G F D
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Traversals
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Printing the Tree
Traversal Algorithm Inorder traversal
What is the order of the output?
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Iterator

• Recall iterator functions for ADT List?
• ResetList()
• GetNextItem()
• Binary Search Tree ResetTree(), GetNextItem()
• Provide different traversal order
• In-order, pre-order, post-order
• Use a parameter to select which traversal order

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Iterator Implementation

• ResetTree generates a queue of node contents in
  the indicated order
• Add private data members three queues storing
  ItemTypes (tree nodes content)
• QueType inQue, preQue, postQue
• GetNextItem returns next node content from the
  appropriate queue

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Iterator

• void TreeTypeResetTree(OrderType order)
• // Calls function to create a queue of the tree
• // elements in the desired order.
• switch (order)
• case PRE_ORDER preQue new QueueType
• PreOrder(root, preQue)
• break
• case IN_ORDER inQue new QueueType
• InOrder(root, inQue)
• break
• case POST_ORDER postQue new QueueType
• PostOrder(root, postQue)
• break

Insert the info field of nodes in tree (given by
root) into preQue in pre-order
What if the queue is not empty when ResetTree()
is called ?
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• void TreeTypeGetNextItem(ItemType item,
• OrderType order, bool finished)
• finished false
• switch (order)
• case PRE_ORDER preQue-gtDequeue(item)
• if (preQue-gtIsEmpty())
• finished true delete
  preQue
• preQue NULL
• break
• case IN_ORDER inQue-gtDequeue(item)
• if (inQue-gtIsEmpty())
• finished true delete
  inQue
• inQue NULL
• break
• case POST_ORDER postQue-gtDequeue(item)
• if (postQue-gtIsEmpty())
• finished true delete
  postQue

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PreOrder() function

• void PreOrder(TreeNode treeRoot, QueType
  preQue)
• if (treeRoot!NULL)
• preQue-gtenqueue (treeRoot-gtinfo)
• PreOrder (treeRoot-gtleft, preQue)
• PreQrder (treeRoot-gtright, preQue)

Hint for recursive functions on a binary tree,
making an empty tree case your base case will
lead to a cleaner and simpler code.
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Printing the Tree

• PrintTree operation
• Size?
• Base case(s)?
• General case(s)?

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Printing the Tree

• void TreeTypePrintTree(TreeNode tree,
  stdofstream outFile)
• if (tree ! NULL)
• PrintTree(tree-gtleft, outFile)
• outFile ltlt tree-gtinfo
• PrintTree(tree-gtright, outFile)

Does this allow a reconstruction of the tree
(same shaped) based on the output?
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Lab6 breadth-first traversal of tree

• Visit the nodes in the order of their layers,
  within same layer, visit nodes from left to
  right
• For the tree in figure
• D B F A C E G
• Printout in this order can be used to reconstruct
  the tree, why?
• Could make tree printout more readable
• D
• / \
• B F
• /\ /\
• A C E G

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BST Destructor

• Do we need a destructor?
• Yes as dynamic allocated memory is used
• delete all nodes (free memory space taken by
  them).
• In which order shall we delete the tree nodes?
• i.e., which Traversal order inorder, preorder,
  postorder?
• TreeTypeTreeType()
• DeleteTree(root)
• root NULL
• void TreeTypeDeleteTreeType(TreeNode root)
• if (root!NULL)
• DeleteTree(root-gtleft)
• DeleteTree(root-gtright)
• delete root

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Copy a Tree

• Copy constructor allow initialization of an
  object by copying from an existing one
• TreeType newTree(oldTree)
• Syntax of a copy constructor
• TreeTypeTreeType(const TreeType
  originalTree)
• Assignment operator overloaded
• TreeType newTree
• newTreeoldTree
• Syntax of assignment operator
• Void TreeTypeoperator(const TreeType
  originalTree)
• Both involves copying a tree
• Use which traversal order?
• Preorder Traversal

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Copying a Tree

• CopyTree(copy, originalTree)
• if (originalTree is NULL)
• Set copy to NULL
• else
• Set copy to new node
• Set Info(copy) to Info(originalTree)
• CopyTree(Left(copy), Left(originalTree))
• CopyTree(Right(copy), Right(originalTree))

How must copy be passed? How must originalTree be
passed?
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Copy a Tree

• void CopyTree (TreeNode copy, const TreeNode
  originalTree)
• if (originalTreeNULL)
• copyNULL
• else
• copy new TreeNode
• copy-gtinfo originalTree-gtinfo
• CopyTree (copy-gtleft, originalTree-gtleft)
• CopyTree (copy-gtright, originalTree-gtright)

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Recursive solutions so far.Now, lets try to
solve some of the problem iteratively.
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Iterative Search

• FindNode(tree, item, nodePtr, parentPtr)
• Searching an item
• If a node is found, get two pointers, one points
  to the node and one points to its parent node.
• If root is the node matching, the first pointers
  points to the root and the second pointer points
  to NULL.
• If no node is found, the first pointers points to
  NULL and the second pointers points to its
  logical parent node.

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Iterative Versions

• FindNode(tree, item, nodePtr, parentPtr)
• Set nodePtr to tree
• Set parentPtr to NULL
• Set found to false
•  
• while more elements to search AND NOT found
• if item lt Info(nodePtr) // search left subtree
• Set parentPtr to nodePtr
• Set nodePtr to Left(nodePtr)
• else if item gt Info(nodePtr) // search right
  subtree
• Set parentPtr to nodePtr
• Set nodePtr to Right(nodePtr)
• else //match
• Set found to true

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• void TreeTypeFindNode(TreeNode tree, ItemType
  item,
• TreeNode nodePtr, TreeNode
  parentPtr)
• nodePtr tree
• parentPtr NULL
• bool found false
• while( nodePtr ! NULL found false)
• if (item.ComparedTo(nodePtr-gtinfo) LESS)
• parentPtr nodePtr
• nodePtr nodePtr-gtleft
• else if (item.ComparedTo(nodePtr-gtinfo)
  LARGER)
• parentPtr nodePtr
• nodePtr nodePtr-gtright
• else
• found true

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Iterative insertion

• InsertItem
• Create a node to contain the new item
• Find the insertion place
• Attach new node

• Find the insertion place
• FindNode(tree, item, nodePtr, parentPtr)
• If no node is found, the first pointers points to
  NULL and the second pointers points to its
  logical parent node.

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Iterative Versions
Insert 13
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Iterative Versions
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Iterative Versions
AttachNewNode if item lt Info(parentPtr) Set
Left(parentPtr) to newNode else Set
Right(parentPtr) to newNode
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Iterative Version

• AttachNewNode(revised)
• if parentPtr equals NULL
• Set tree to newNode
• else if item lt Info(parentPtr)
• Set Left(parentPtr) to newNode
• else
• Set Right(parentPtr) to newNode

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Iterative Version DeleteItem

• void TreeTypeDeleteItem(ItemType item)
• TreeNode nodePtr
• TreeNode parentPtr
• FindNode(root, item, nodePtr, parentPtr)
• if (nodePtr root)
• DeleteNode(root)
• else
• if (parentPtr-gtleft nodePtr)
• DeleteNode(parentPtr-gtleft)
• else DeleteNode(parentPtr-gtright)

Why not directly delete nodePtr?
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Recursive Deletion review

• void TreeTypeDeleteNode(TreeNode tree)
• ItemType data
• TreeNode tempPtr
• tempPtr tree
• if ( tree-gtleft NULL)
• tree tree-gtright
• delete tempPtr
• else if (tree-gtright NULL)
• tree tree-gtleft
• delete tempPtr
• else
• GetPredecessor(tree-gtleft, data)
• tree-gtinfo data
• Delete(tree-gtleft, data)
  

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Iterative Version
parentPtr and nodePtr are external
parentPtr-gtleft is internal (to the tree)
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Recursion vs. Iteration
Compare versions of Tree algorithms Is the depth
of recursion relatively shallow? Is the
recursive solution shorter or cleaner? Is the
recursive version much less efficient?
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Nonlinked Representation
What is the mapping into the index?
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Nonlinked Representation

• For any node tree.nodesindex
• its left child is in tree.nodesindex2 1
• right child is in tree.nodesindex2 2
• its parent is in tree.nodes(index - 1)/2
• Can you determine which nodes are leaf nodes?

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Nonlinked Representation
Full Binary Tree A binary tree in which all of
the leaves are on the same level and every
nonleaf node has two children
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Nonlinked Representation
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
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Nonlinked Representation
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Nonlinked Representation
A technique for storing a non-complete tree
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Binary Search Trees
The same set of keys may have different BSTs

• Average depth of a node is O(log2 N)
• Maximum depth of a node is O(N)

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Time Complexity

• Time complexity of Searching
• O(height of the tree)
• Time complexity of Inserting
• O(height of the tree)
• Time complexity of Deleting
• O(height of the tree)

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Comparison

• Assume that we have a Complete tree.

Binary Search Tree Array-based Linear List Linked List
Constructor O(1) O(1) O(1)
Destructor O(N) O(1) (non-pointers) O(N)
MakeEmpty O(N) O(1) O(N)
GetLength O(N) O(1) O(1)
IsFull O(1) O(1) O(1)
IsEmpty O(1) O(1) O(1)
RetrieveItem O(log2N) O(log2N) O(N)
InsertItem O(log2N) O(N) O(N)
DeleteItem O(log2N) O(N) O(N)
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Reference

• Reproduced from C Plus Data Structures, 4th
  edition by Nell Dale.
• Reproduced by permission of Jones and Bartlett
  Publishers International.