Binary Search Tree Algorithm

Binary Search Tree Algorithm 

Suppose T is a binary tree. Then, T is called a binary search tree (or binary sorted tree) if each node N of T has the following property:

The value at N is greater than every value in the left subtree of N and is less than every value in the right subtree of N.

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Searching and Inserting in Binary search tree

The searching and inserting will be given by a single search and insertion algorithm.

Suppose an ITEM of information is given. The following algorithm finds the location of ITEM in the binary search tree T, or inserts ITEM as a new node in its appropriate place in the tree.

a)      Compare ITEM with the root node N of the tree.

(i)     If ITEM < N, proceed to the left child of N.

(ii)   If ITEM > N, proceed to the right child of N.

b)      Repeat step (a) until one of the following occurs:

(i)     We meet a node N such that ITEM = N. in this case the search is successful.

(ii)   We meet an empty subtree, which indicate that the search is unsuccessful, and we insert ITEM in place of the empty subtree.

In other words, proceed from the root R down through the tree T until finding ITEM in T or inserting ITEM as a terminal node in T.

The formal presentation of our search and insertion algorithm will use the following procedure, which finds the location of a given ITEM and its parent. The procedure traverses down the tree using the pointer PTR and the pointer SAVE for the parent node. This procedure will also be used in the next section, or deletion.

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Searching

Algorithm: FIND(INFO, LEFT, RIGHT, ROOT, ITEM, LOC, PAR)

A binary tree T is in memory and an ITEM of information is given. This procedure finds the location LOC of ITEM in T and also the location PAR of the parent of ITEM. There are three special cases:

(i)     LOC = NULL and PAR = NULL will indicate that the tree is empty. [No node.]

(ii)   LOC != NULL and PAR = NULL will indicate that ITEM is root of T. [No parent of node, i.e., only one node.]

(iii)  LOC = NULL and PAR != NULL will indicate that ITEM is not in T and can be added to T as a child of the node N with the location PAR.

Step I:              [Tree empty?]

If ROOT = NULL, then:

Set LOC := NULL and PAR := NULL, and Return.

Step II:           [ITEM at root?]

If ITEM = INFO[ROOT], then:

Set LOC := ROOT and PAR := NULL, and Return.

Step III:        [Initialize pointers PTR and SAVE.]

If ITEM < INFO[ROOT], then:

Set SAVE := ROOT and PTR := LEFT[ROOT]

Else:

Set SAVE := ROOT and PTR := RIGHT[ROOT]

[End of If structure.]

Step IV:        Repeat steps 5 and 6 while PTR != NULL;

Step V:           [ITEM found?]

If ITEM := INFO[PTR], then:

Set LOC := PTR and PAR := SAVE, and Return.

Step VI:        If ITEM < INFO[PTR], then:

Set SAVE := PTR and PTR := LEFT[PTR].

Else:

Set SAVE := PTR and PTR := RIGHT[PTR].

[End of If structure.]

[End of step 4 loop.]

Step VII:     [Search unsuccessful.]

Set LOC := NULL and PAR := SAVE.

Step VIII:  Return.

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