Trees, Binary Trees, and Binary Search Trees. 2 Trees * Linear access time of linked lists is...

Trees, Binary Trees, and Binary Search Trees

2

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

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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

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Tree can be viewed as a ‘nested’ lists Tree is also a graph …

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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

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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

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Example: UNIX Directory

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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

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Example: Unix Directory TraversalPreOrder PostOrder

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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

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Convert a Generic Tree to a Binary Tree

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Binary Tree ADT

Possible operations on the Binary Tree ADT Parent, left_child, right_child, sibling, root, etc

Implementation Because a binary tree has at most two children, we can keep

direct pointers to them a linked list is physically a pointer, so is a tree.

Define a Binary Tree ADT later …

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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

}

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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

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Preorder, Postorder and Inorder

Preorder traversal node, left, right prefix expression

++a*bc*+*defg

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Preorder, Postorder and Inorder

Postorder traversal left, right, node postfix expression

abc*+de*f+g*+

Inorder traversal left, node, right infix expression

a+b*c+d*e+f*g

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Preorder, Postorder and Inorder Pseudo Code

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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

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Binary Search Trees

A binary search tree Not a binary search tree

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Binary Search Trees

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

The same set of keys may have different BSTs

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Searching BST If we are searching for 15, then we are done. If we are searching for a key < 15, then we

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

should search in the right subtree.

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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

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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

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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

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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)

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void insert(double x, BinaryNode*& t)

{

if (t==NULL) t = new BinaryNode(x,NULL,NULL);

else if (x<t->element) insert(x,t->left);

else if (t->element<x) insert(x,t->right);

else ; // do nothing

}

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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.

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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

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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)

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void remove(double x, BinaryNode*& t)

{

if (t==NULL) return;

if (x<t->element) remove(x,t->left);

else if (t->element < x) remove (x, t->right);

else if (t->left != NULL && t->right != NULL) // two children

{

t->element = finMin(t->right) ->element;

remove(t->element,t->right);

}

else

{

Binarynode* oldNode = t;

t = (t->left != NULL) ? t->left : t->right;

delete oldNode;

}

}

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Insertion Example Construct a BST successively from a sequence of data: 35,60,2,80,40,85,32,33,31,5,30

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Deletion Example Removing 40 from (a) results in (b) using the smallest

element in the right subtree (i.e. the successor)

5

30

2

40

8035

32

3331

8560

5

30

2

60

8035

32

3331

85

(a) (b)

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Removing 40 from (a) results in (c) using the largest element in the left subtree (i.e., the predecessor)

COMP152 34

5

30

2

40

8035

32

3331

8560

(a) (c)

5

30

2

35

8032

3331 8560

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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.

COMP152 35

(c)

5

30

2

35

8032

3331 8560

(d)

2

5

35

8032

3331 8560

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Example: compute the number of nodes?

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Search trees 37

Example: SuccessorThe 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)

Successor of 2Successor of 6

Successor of 13

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

Successor of 9

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Search trees 38

Finding Successor:Three Scenarios to Determine Successor

Successor(x)

x has right descendants=> minimum( right(x) )

x has no right descendants

x is the left child of some node=> parent(x)

x is the right child of some node

Scenario I

Scenario II Scenario III

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Search trees 39

Scenario I: Node x Has a Right Subtree

By definition of BST, all items greater thanx are in this right sub-tree.

Successor is the minimum( right( x ) )

maybe null

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Search trees 40

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 whosekey would appear in the next sorted order.

Think about traversal in-order. Who wouldbe the successor of x?

The parent of x!

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Search trees 41

Scenario III: Node x Has No Right Subtree and Is Not a Left-Child of an

Immediate Parent

Keep moving up the tree untilyou find a parent which branchesfrom the left().Successor of x

Stated in Pseudo code.y

x

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Search trees 42

Successor Pseudo-Codes

Scenario I

Scenario III

Verify this codewith this tree.

Find successor of 3 49 1313 1518 20

Scenario II

Note that parent( root ) = NULL

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Search trees 43

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 use a Stack.

class Node{

int data;Node * left;Node * right;Node *

parent;};

class Node{

int data;Node * left;Node *

right;};

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Search trees 44

Use a Stack to Find SuccessorPART IInitialize 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 IIOnce node x is found, find successorusing 3 scenarios mentioned before.

Parent nodes are found by popping the stack!

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Search trees 45

An Example

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6

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Stack s

Successor(root, 13)Part I Traverse tree from root to find 13 order -> 15, 6, 7, 13

push(15)

push(6)

push(7)

13 found (x = node 13)

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Search trees 46

Example

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6

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Stack s

Successor(root, 13)Part II Find Parent (Scenario III) y=s.pop() while y!=NULL and x=right(y) x = y; if s.isempty() y=NULL else y=s.pop() loop return y

y =pop()=15 ->Stop right(15) != x return y as successor!

x = 13

y =pop()=7

y =pop()=6

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Make a binary or BST ADT …

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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;

}

For a generic (binary) tree:

(insert and remove are different from those of BST)

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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;

}

For BST tree:

BST is for efficient search, insertion and removal, so restricting these functions.

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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;

}

Weiss textbook:

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root, left subtree, right subtree are missing:

1. we can’t 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 it’s two layers, the public for BST, and the private for Binary Tree.

3. it might be defined internally in ‘private’ part (actually it’s implicitly done).

Comments:

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void insert(double x, BinaryNode*& t)

{

if (t==NULL) t = new BinaryNode(x,NULL,NULL);

else if (x<t->element) insert(x,t->left);

else if (t->element<x) insert(x,t->right);

else ; // do nothing

}

void insert(double x)

{

insert(x,root);

}

A public non-recursive member function:

A private recursive member function:

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Struct Node {

double element; // the data

Node* left; // left child

Node* right; // right child

}

Class BinaryTree {

…

}

class BST : public BinaryTree {

}

void BST::search () {

}

void BST::insert () {

}

void BST::delete () {

}

By inheritance

All search, insert and deletion have to be redefined.

template<typename T>

Struct Node {

T element; // the data

Node* left; // left child

Node* right; // right child

}

template<typename T>

class BinaryTree {

…

}

template<typename T>

class BST : public BinaryTree<T> {

}

void BST<T>::search (const T& x) {

}

void BST<T>::insert (…) {

}

void BST<T>::delete (…) {

}

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More general BST …template<typename T>

Struct Node {

T element; // the data

Node* left; // left child

Node* right; // right child

}

template<typename T>

class BinaryTree {

…

}

template<typename T>

class BST : public BinaryTree<T> {

}

void BST<T>::search (const T& x) {

}

void BST<T>::insert (…) {

}

void BST<T>::delete (…) {

}

template<typename T>

class BinaryTree {

…

}

template<typename T, typename K>

class BST : public BinaryTree<T> {

}

void BST<T>::search (const K& key) {

}

Search ‘key’ of K might be different from the data record of T!!!

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Deletion Code (1/4) First Element Search, and then Convert Case III, if any,

to Case I or II

COMP152 55

template<class E, class K>BSTree<E,K>& BSTree<E,K>::Delete(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)BinaryTreeNode<E> *p = root, // search pointer

*pp = 0; // parent of pwhile (p && p->data != k){

// move to a child of ppp = p;if (k < p->data) p = p->LeftChild;else p = p->RightChild;

}

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Deletion Code (2/4)

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

e = p->data; // save element to delete

// restructure tree// handle case when p has two childrenif (p->LeftChild && p->RightChild) {

// two children convert to zero or one child case// find predecessor, i.e., the largest element in // left subtree of pBinaryTreeNode<E> *s = p->LeftChild,*ps = p; // parent of swhile (s->RightChild) {

// move to larger elementps = s;s = s->RightChild;}

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Deletion Code (3/4)

COMP152 57

// move from s to pp->data = s->data;p = s; // move/reposition pointers for deletionpp = ps;

}

// p now has at most one child// save child pointer to c for adoptionBinaryTreeNode<E> *c;if (p->LeftChild) c = p->LeftChild;else c = p->RightChild;

// deleting pif (p == root) root = c; // a special case: delete rootelse {

// is p left or right child of pp?if (p == pp->LeftChild) pp->LeftChild = c;//adoptionelse pp->RightChild = c;}

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Deletion Code (4/4)

COMP152 58

delete p;

return *this;}