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Trees, Binary Trees, and Binary Search Trees

Trees

- Linear access time of linked lists is prohibitive
- Does there exist any simple data structure for

which the running time of most operations

(search, insert, delete) is O(log N)? - Trees
- Basic concepts
- Tree traversal
- Binary tree
- Binary search tree and its operations

Trees

- A tree T is a collection of nodes
- T can be empty
- (recursive definition) If not empty, a tree T

consists of - a (distinguished) node r (the root),
- and zero or more nonempty sub-trees T1, T2,

...., Tk

- Tree can be viewed as a nested lists
- list of lists of lists
- Tree is also a graph

Some Terminologies

- Child and Parent
- Every node except the root has one parent
- A node can have a zero or more children
- Leaves
- Leaves are nodes with no children
- Sibling
- nodes with same parent

More Terminologies

- Path
- A sequence of edges
- Length of a path
- number of edges on the path
- Depth of a node
- length of the unique path from the root to that

node - Height of a node
- length of the longest path from that node to a

leaf - all leaves are at height 0
- The height of a tree the height of the root

the depth of the deepest leaf - Ancestor and descendant
- If there is a path from n1 to n2
- n1 is an ancestor of n2, n2 is a descendant of n1
- Proper ancestor and proper descendant

Example UNIX Directory

Tree Operations

- Traversal, the most important
- we will not implement a general tree, so wont

discuss - Search
- Insertion
- Deletion

Tree Traversal

- Used to print out the data in a tree in a certain

order - Pre-order traversal
- Print the data at the root
- Recursively print out all data in the leftmost

subtree - Recursively print out all data in the rightmost

subtree

A drawing of tree with two pointers

Struct TreeNode double element // the data

TreeNode child // FIRST child go to the

next generation TreeNode next // next SIBLING

go to the same generation

preorder(tree) if empty(tree) stop else

print(element(tree)) preorder(child(tree))

preorder(next(tree))

Example Unix Directory Traversal

PreOrder

PostOrder

Binary Trees

- A tree in which no node can have more than two

children - The depth of an average binary tree is

considerably smaller than N, even though in the

worst case, the depth can be as large as N 1.

Generic binary tree

Worst-casebinary tree

Binary Tree ADT

- Implementation
- at most two children, we can keep direct pointers

to them - a linked list is physically a pointer, so is a

tree! - Possible operations on the Binary Tree ADT
- Parent, left_child, right_child, sibling, root,

etc - Tree operations
- Search, insertion and deletion
- Make a Binary Tree ADT later

A drawing of linked list with one pointer

A drawing of binary tree with two pointers

Struct BinaryNode double element // the data

BinaryNode left // left child BinaryNode

right // right child

Example Expression Trees

- Leaves are operands (constants or variables)
- The internal nodes contain operators
- Will not be a binary tree if some operators are

not binary

Preorder, Postorder and Inorder

- Preorder traversal
- node, left, right
- prefix expression
- abcdefg

Preorder, Postorder and Inorder

- Inorder traversal
- left, node, right
- infix expression
- abcdefg

- Postorder traversal
- left, right, node
- postfix expression
- abcdefg

Preorder, Postorder and Inorder Pseudo Code

Binary tree insertion and deletion

BST Binary Search Tree

Recall the binary search algorithm

Binary Search Trees (BST)

- A data structure for efficient searching,

inser-tion and deletion - Binary search tree property
- For every node X
- All the keys in its left subtree are smaller

than the key value in X - All the keys in its right subtree are larger

than the key value in X

Binary Search Trees

A binary search tree

Not a binary search tree

Binary Search Trees

The same set of keys may have different BSTs

- Average depth of a node is O(log N)
- Maximum depth of a node is O(N)

Searching BST

- If we are searching for 15, then we are done.
- If we are searching for a key lt 15, then we

should search in the left subtree. - If we are searching for a key gt 15, then we

should search in the right subtree.

(No Transcript)

Searching (Find)

- Find X return a pointer to the node that has key

X, or NULL if there is no such node - Time complexity O(height of the tree)

find(const double x, BinaryNode t) const

Inorder Traversal of BST

- Inorder traversal of BST prints out all the keys

in sorted order

Inorder 2, 3, 4, 6, 7, 9, 13, 15, 17, 18, 20

findMin/ findMax

- Goal return the node containing the smallest

(largest) key in the tree - Algorithm Start at the root and go left (right)

as long as there is a left (right) child. The

stopping point is the smallest (largest) element - Time complexity O(height of the tree)

BinaryNode findMin(BinaryNode t) const

Insertion

- Proceed down the tree as you would with a find
- If X is found, do nothing (or update something)
- Otherwise, insert X at the last spot on the path

traversed - Time complexity O(height of the tree)

void insert(double x, BinaryNode t) if

(tNULL) t new BinaryNode(x,NULL,NULL) else

if (xltt-gtelement) insert(x,t-gtleft) else if

(t-gtelementltx) insert(x,t-gtright) else // do

nothing

Insertion Example

- Construct a BST successively from a sequence of

data - 35,60,2,80,40,85,32,33,31,5,30

Deletion

- When we delete a node, we need to consider how we

take care of the children of the deleted node. - This has to be done such that the property of the

search tree is maintained.

Deletion under Different Cases

- Case 1 the node is a leaf
- Delete it immediately
- Case 2 the node has one child
- Adjust a pointer from the parent to bypass that

node

Deletion Case 3

- Case 3 the node has 2 children
- Replace the key of that node with the minimum

element at the right subtree - (or replace the key by the maximum at the left

subtree!) - Delete that minimum element
- Has either no child or only right child because

if it has a left child, that left child would be

smaller and would have been chosen. So invoke

case 1 or 2. - Time complexity O(height of the tree)

void remove(double x, BinaryNode t) if

(tNULL) return if (xltt-gtelement)

remove(x,t-gtleft) else if (t-gtelement lt x)

remove (x, t-gtright) else if (t-gtleft ! NULL

t-gtright ! NULL) // two children t-gteleme

nt finMin(t-gtright) -gtelement remove(t-gteleme

nt,t-gtright) else Binarynode oldNode

t t (t-gtleft ! NULL) ? t-gtleft

t-gtright delete oldNode

Deletion Example

- Removing 40 from (a) results in (b) using the

smallest element in the right subtree (i.e. the

successor)

(b)

(a)

- Removing 40 from (a) results in (c) using the

largest element in the left subtree (i.e., the

predecessor)

(c)

(a)

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- Removing 30 from (c), we may replace the element

with either 5 (predecessor) or 31 (successor). If

we choose 5, then (d) results.

(d)

(c)

40

Example compute the number of nodes?

Example Successor

- The successor of a node x is
- defined as
- The node y, whose key(y) is the successor of

key(x) in sorted order - sorted order of this tree.

(2,3,4,6,7,9,13,15,17,18,20)

Some examples Which node is the successor of

2? Which node is the successor of 9? Which node

is the successor of 13? Which node is the

successor of 20? Null

Search trees

42

Scenario I Node x Has a Right Subtree

By definition of BST, all items greater than x

are in this right sub-tree. Successor is the

minimum( right( x ) )

maybe null

Search trees

43

Scenario II Node x Has No Right Subtree and x is

the Left Child of Parent (x)

Successor is parent( x ) Why? The successor is

the node whose key would appear in the next

sorted order. Think about traversal in-order.

Who wouldbe the successor of x? The parent of

x!

Search trees

44

Scenario III Node x Has No Right Subtree and Is

Not a Left-Child of an Immediate Parent

Keep moving up the tree until you find a parent

which branches from the left().

Successor of x

y

Stated in Pseudo code.

x

Search trees

45

Three Scenarios to Determine Successor

Successor(x)

x has right descendants gt minimum( right(x) )

x has no right descendants

Scenario I

x is the left child of some node gt parent(x)

x is the right child of some node

Scenario II

Scenario III

Search trees

46

Successor Pseudo-Codes

Verify this code with this tree. Find successor

of 3 ? 4 9 ? 13 13 ? 15 18 ? 20

Note that parent( root ) NULL

Scenario I

Scenario II

Scenario III

Search trees

47

Problem

- If we use a doubly linked tree, finding parent

is easy. - But usually, we implement the tree using only

pointers to the left and right node. ? So,

finding the parent is tricky. - For this implementation we need to store the

path ? Stack!

class Node int data Node left Node

right Node parent

class Node int data Node left Node

right

48

Use a Stack to Find Successor

PART I Initialize an empty Stack s. Start at the

root node, and traverse the tree until we find

the node x. Push all visited nodes onto the

stack.

PART II Once node x is found, find

successor using 3 scenarios mentioned

before. Parent nodes are found by popping the

stack!

49

Example

Successor(root, 13) Part I Traverse tree from

root to find 13 order -gt 15, 6, 7, 13

7

6

15

Stack s

50

Successor(root, 13) Part II Find Parent

(Scenario III) ys.pop() while y!NULL

and xright(y) x y if s.isempty()

yNULL else ys.pop() return

y

y pop()6

y pop()7

x 13

7

6

15

Stack s

51

Make a binary or BST ADT

For a generic (binary) tree

Struct Node double element // the data

Node left // left child Node right //

right child class Tree public Tree()

//

constructor Tree(const Tree t) Tree()

//

destructor bool empty() const double root()

// decomposition (access functions) Tree

left() Tree right() bool search(const

double x) void insert(const double x) //

compose x into a tree void remove(const double

x) // decompose x from a tree private Node

root

(insert and remove are different from those of

BST)

For BST tree

Struct Node double element // the data

Node left // left child Node right //

right child class BST public BST()

//

constructor BST(const Tree t) BST()

//

destructor bool empty() const double root()

// decomposition (access functions) BST

left() BST right() bool serch(const double

x) // search an element void insert(const

double x) // compose x into a tree void

remove(const double x) // decompose x from a

tree private Node root

BST is for efficient search, insertion and

removal, so restricting these functions.

Weiss textbook

class BST public BST() BST(const Tree

t) BST() bool empty() const bool

search(const double x) // contains void

insert(const double x) // compose x into a

tree void remove(const double x) // decompose x

from a tree private Struct Node double

element Node left Node right Node()

// constructuro for Node Node

root void insert(const double x, Node t)

const // recursive function void

remove() Node findMin(Node t) void

makeEmpty(Node t) // recursive

destructor bool contains(const double x, Node

t) const

Comments

root, left subtree, right subtree are

missing 1. we cant write other tree

algorithms, is implementation dependent,

BUT, 2. this is only for BST (we only need

search, insert and remove, may not need other

tree algorithms) so its two layers, the

public for BST, and the private for Binary

Tree. 3. it might be defined internally in

private part (actually its implicitly done).

A public non-recursive member function

void insert(double x) insert(x,root)

A private recursive member function

void insert(double x, BinaryNode t) if

(tNULL) t new BinaryNode(x,NULL,NULL) else

if (xltt-gtelement) insert(x,t-gtleft) else if

(t-gtelementltx) insert(x,t-gtright) else // do

nothing

By inheritance

Struct Node double element // the data

Node left // left child Node right //

right child Class BinaryTree class BST

public BinaryTree void BSTsearch ()

void BSTinsert () void BSTdelete ()

templatelttypename Tgt Struct Node T element //

the data Node left // left child Node

right // right child templatelttypename

Tgt class BinaryTree templatelttypename

Tgt class BST public BinaryTreeltTgt void

BSTltTgtsearch (const T x) void

BSTltTgtinsert () void BSTltTgtdelete ()

All search, insert and deletion have to be

redefined.

More general BST

templatelttypename Tgt Struct Node T element //

the data Node left // left child Node

right // right child templatelttypename

Tgt class BinaryTree templatelttypename

Tgt class BST public BinaryTreeltTgt void

BSTltTgtsearch (const T x) void

BSTltTgtinsert () void BSTltTgtdelete ()

templatelttypename Tgt class BinaryTree

templatelttypename T, typename Kgt class

BST public BinaryTreeltTgt void

BSTltTgtsearch (const K key)

Search key of K might be different from the

data record of T!!!

Deletion Code (1/4)

- First Element Search, and then Convert Case III,

if any, to Case I or II

templateltclass E, class Kgt BSTreeltE,Kgt

BSTreeltE,KgtDelete(const K k, E e) // Delete

element with key k and put it in e. // set p to

point to node with key k (to be

deleted) BinaryTreeNodeltEgt p root, // search

pointer pp 0 // parent of p while (p

p-gtdata ! k) // move to a child of p pp

p if (k lt p-gtdata) p p-gtLeftChild else p

p-gtRightChild

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Deletion Code (2/4)

if (!p) throw BadInput() // no element with key

k e p-gtdata // save element to delete //

restructure tree // handle case when p has two

children if (p-gtLeftChild p-gtRightChild) //

two children convert to zero or one child case //

find predecessor, i.e., the largest element in //

left subtree of p BinaryTreeNodeltEgt s

p-gtLeftChild, ps p // parent of s while

(s-gtRightChild) // move to larger element ps

s s s-gtRightChild

61

Deletion Code (3/4)

// move from s to p p-gtdata s-gtdata p s

// move/reposition pointers for deletion pp

ps // p now has at most one child // save

child pointer to c for adoption BinaryTreeNodeltEgt

c if (p-gtLeftChild) c p-gtLeftChild else c

p-gtRightChild // deleting p if (p root)

root c // a special case delete root else

// is p left or right child of pp? if (p

pp-gtLeftChild) pp-gtLeftChild c//adoption else

pp-gtRightChild c

62

Deletion Code (4/4)

delete p return this

63