**Q. Write down an algorithm to delete the specific node from binary search tree. Trace the algorithm to delete a node (10) from the following given tree.**

**Ans.**

**Algorithm for Delete ting the specific Node From the Binary Search Tree**

To delete the specific node following possibilities may arise

1) Node id a terminal node

2) Node have only one child

3) Node having 2 children.

DEL(INFO, LEFT, RIGT, ROOT, AVAIL, ITEM)

A binary search tree T is in the memory, and an ITEM of information is given as follows.

This algorithm deletes the specific ITEM from the tree.

1. [to Find the locations of ITEM and its parent] Call FIND(INFO, RIGHT, ROOT, ITEM, LOC, PAR).

2. [ITEM in tree?]

if LOC=NULL, then write : ITEM not in tree, and Exit.

3. [Delete node containing ITEM.]

if RIGHT[LOC] != NULL and LEFT[LOC] !=NULL then:

Call CASEB(INFO,LEFT,RIGHT,ROOT,LOC,PAR). Else:

Call CASEA (INFO,LEFT,RIGHT,ROOT,LOC,PAR).

[End of if structure.]

4. [Return deleted node to AVAIL list.] Set LEFT[LOC]:=AVAIL and AVAIL:=LOC.

5. Exit.

CASEB(INFO,LEFT,RIGHT,ROOT,LOC,PAR)

This procedure will delete the node N at LOC location, where N has two children. The pointer PAR gives us the location of the parent of N, or else PAR=NULL indicates that N is a root node. The pointer SUC gives us the location of the inorder successor of N, and PARSUC gives us the location of the parent of the inorder successor.

1. [Find SUC and PARSUC.]

(a) Set PTR: = RIGHT[LOC] and SAVE:=LOC. (b) Repeat while LEFT[PTR] ≠ NULL:

Set SAVE:=PTR and PTR:=LEFT[PTR]. [End of loop.]

(c) Set SUC : = PTR and PARSUC:=SAVE.

2. [Delete inorder successor]

Call CASEA (INFO, LEFT, RIGHT, ROOT, SUC, PARSUC).

3. [Replace node N by its inorder successor.] (a) If PAR≠NULL, then:

If LOC = LEFT[PAR], then: Set LEFT[PAR]:=SUC.

Else:

Set RIGHT[PAR]: = SUC. [End of If structure.]

Else:

Set ROOT: = SUC. [End of If structure.]

(b) Set LEFT[SUC]:= LEFT [LOC] and

RIGHT[SUC]:=RIGHT[LOC]

4. Return.

CASEA(INFO, LEFT, RIGHT, ROOT, LOC, PAR)

This procedure deletes the node N at LOC location, where N does not contain two children. The pointer PAR gives us the location of the parent of N, or else PAR=NULL indicates that N is a root node. The pointer CHILD gives us the location of the only child of the N, or else CHILD = NULL indicates N has no children.

1. [Initializes CHILD.]

If LEFT[LOC] = NULL and RIGHT[LOC] = NULL, then: Set CHILD:=NULL.

Else if LEFT[LOC]≠NULL, then:

Set CHILD: = LEFT[LOC].

Else

Set CHILD:=RIGHT[LOC] [End of If structue.]

2. If PAR ≠ NULL, then:

If LOC = LEFT [PAR], then:

Set LEFT[PAR]:=CHILD.

Else:

Set RIGHT[PAR]:CHILD = CHILD [End of If structure.]

Else:

Set ROOT : = CHILD.

[End of If structure.]

3. Return.

Inorder traversal of the tree is

4 6 10 11 12 14 15 20

To delete 10

PAR = Parent of 10 ie 15

SUC = inorder succ of 10 ie. 11

PARSUC = Parent of inorder succ ie 12

PTR = RIGHT [LOC]

Address of 12 SAVE: = address of 10

SAVE: = address of 12

PTR = address of 11

SUC = ADDRESS OF 11

PAR SUCC:= ADDRESS OF 12

CHILD = NULL

LEFT [PARSUC] = CHILD= NULL LEFT [PAR]= ADDRESS OF 11

LEFT [SUC] = LEFT [LOC] = ADDRESS OF 6

RIGHT [SUC] = RIGHT[LOC] = ADDRESS OF 12