Non-Linear Data Structure

Binary Trees

Non-Linear Data Structure

• These data structures donot have their elements in a sequence.

• Trees is an example.

• Trees are mainly used to represent data containing a hierarchical relationship between elements, ex : records, family trees and table of contents.

Tree example

Trees

• Tree nodes contain two or more links

– All other data structures we have discussed

only contain one

• Binary trees

– All nodes contain two links

• None, one, or both of which may be NULL

– The root node is the first node in a tree.

– Each link in the root node refers to a child

– A node with no children is called a leaf node

Binary Trees

• Special type of tree in which every node or vertex has either no children, one child or two children.

• Characteristics :• Every binary tree has a root pointer which

points to the start of the tree.

• A binary tree can be empty.

• It consists of a node called root, a left subtreeand right subtree both of which are binary trees themselves.

Examples : Binary Trees

X X X X

Y YZ Z

A

B C

(1) (2) (3) (4)

Root of tree is node having info as X.

(1) Only node is root.

(2) Root has left child Y.

(3) Root X has right child Z.

(4) Root X has left child Y and right child Z which

is again a binary tree with its parent as Z and

left child of Z is A, which in turn is parent for

left child B and right child C.

Properties of Binary Tree

• A tree with n nodes has exactly (n-1) edges or

branches.

• In a tree every node except the root has exactly

one parent (and the root node does not have a

parent).

• There is exactly one path connecting any two

nodes in a tree.

• The maximum number of nodes in a binary tree

of height K is 2K+1 -1 where K>=0.

Representation of Binary Tree

• Array representation

– The root of the tree is

stored in position 0.

– The node in position p, is

the implicit father of nodes

2p+1 and 2p+2.

– Left child is at 2p+1 and

right at 2p+2.

X

Y Z

A

B C

Y AZ

0 2 3 4 5 6 71 8

B C

10 11 12

X

9

Representation of Binary Tree

• Linked List• Every node will consists of information, and two

pointers left and right pointing to the left and right child nodes.

struct node{

int data;

struct node *left;

struct node *right;

};

• The topmost node or first node is pointed by a root pointer which will inform the start of the tree. Rest of the nodes are attached either to left if less than parent or right if more or equal to parent.

• Diagram of a binary tree

B

A D

C

Operations on Binary Tree

• Searching an existing node.

• Inserting a new node.

• Deleting an existing node.

• Traversing the tree.

• Preorder

• Inorder

• Postorder

Search Process

• Initialize a search pointer as the root pointer.

• All the data is compared with the data stored in each

node of the tree starting from root node.

• If the data to be searched is equal to the data of the

node then print “successful” search.

• Else if the data is less than node’s data move the pointer

to left subtree. Else data is more than node’s data move

the pointer to right subtree.

• Keep moving in the tree until the data is found or search

pointer comes to NULL in which case it is an “unsuccessful” search.

Example used for Search21

18

197

6 9

8 11

14

13

Root

Example : Search

• Initialize temp as root pointer which is node having 21.

• Loop until temp!=NULL or ITEM found

• Compare item=9 with the root 21 of the tree, since 9<21, proceed temp to the left child of the tree.

• Compare ITEM=9 with 18 since 9<18 proceed temp to the left subtree of 18, which is 7.

• Compare ITEM=9 with 7 since 9>7 proceed temp to

right subtree of the node which holds a value 7.

• Now ITEM=9 is compared with the node which holds

value 9 and since 9 is found the searching process ens here.

Insert Process

• For node insertion in a binary search tree, initially the data that is to be inserted is compared with the data of

the root data.

• If the data is found to be greater than or equal to the

data of the root node then the new node is inserted in

the right subtree of the root node, else left subtree.

• Now the parent of the right or left subtree is considered

and its data is compared with the data of the new node and the same procedure is repeated again until a NULL

is found which will indicate a space when the new node has to be attached. Thus finally the new node is made

the appropriate child of this current node.

Example used for Insert38

14

238

Root

56

8245

18 70

20

Example : Insert

• Suppose the ITEM=20 is the data part of the new node to be inserted in the tree and the root is pointing to the

start of the tree.

• Compare ITEM=20 with the root 38 of the tree. Since 20

< 38 proceed to the left child of 38, which is 14.

• Compare ITEM=20 with 14, since 20 > 14 proceed to the right child of 14, which is 23.

• Compare ITEM=20 with 23 since 20 < 23 proceed to the left child of 23 which is 18.

• Compare ITEM=20 with 18 since 20 > 18 and 18 does not have right child, 10 is inserted as the right child of 18.

Deletion

• Condition (i) Print message data not found.

• Condition (ii) Since no children so free the node

& make either parent->left=NULL or parent-

>right=NULL.

A A

B C

Parent Parent

left right

X X

In the example1,

parent->left==x so free(x) and

make parent->left=NULL.

In the example2,

parent->right==x so free(x) and

make parent->right=NULL.

Deletion• Condition (iii) Adjust parent to point to the child of the deleted node.

I. Node deleted(X) has only left child :

1) If(parent->left==x) parent->left=x->left.

2) If(parent->right==x) parent->right=x->left.

A A

B

Parent Parent

left right

X XB

C C

leftright

Deletion• Condition (iii) Adjust parent to point to the child of the deleted node.

II. Node deleted(X) has only right child :

1) If(parent->left==x) parent->left=x->right.

2) If(parent->right==x) parent->right=x->right.

A A

B

Parent Parent

left right

X XB

C C

rightright

Deletion• Condition (iv) solution more complex as X has two children,• Find inorder successor of the node X. Inorder successor of any node will be first go to right of the

node and then from that node keep going left until NULL encountered, that will be the inordersuccessor.

• Copy data of inorder successor to node X’s data.• Place pointer at the inorder successor node. Now the inorder succesor will have zero or one child

only (so the complex problem is reduced to condition (iii)).• Delete the inorder successor node by using logic of condition (ii) if no children or condition (iiii) if one

right child.

A

B

Parent

left

X

D

rightleft

C

E

left

A

E

Parent

left

X

D

rightleft

C

E

leftHere inorder successor E has no children condition (ii)