Question

Create an array-based implementation of a binary tree. (WRITE IN JAVA)

DON'T FORGET TO INCLUDE PSEUDOCODE AND UML DIAGRAM

Answer 1

// JAVA implementation of tree using array
// numbering starting from 0 to n-1.
import java.util.*;
import java.lang.*;
import java.io.*;

class Tree {
        public static void main(String[] args)
        {
                Array_imp obj = new Array_imp();
                obj.Root("A");
                // obj.set_Left("B", 0);
                obj.set_Right("C", 0);
                obj.set_Left("D", 1);
                obj.set_Right("E", 1);
                obj.set_Left("F", 2);
                obj.print_Tree();
        }
}

class Array_imp {
        static int root = 0;
        static String[] str = new String[10];

        /*create root*/
        public void Root(String key)
        {
                str[0] = key;
        }

        /*create left son of root*/
        public void set_Left(String key, int root)
        {
                int t = (root * 2) + 1;

                if (str[root] == null) {
                        System.out.printf("Can't set child at %d, no parent found\n", t);
                }
                else {
                        str[t] = key;
                }
        }

        /*create right son of root*/
        public void set_Right(String key, int root)
        {
                int t = (root * 2) + 2;

                if (str[root] == null) {
                        System.out.printf("Can't set child at %d, no parent found\n", t);
                }
                else {
                        str[t] = key;
                }
        }

        public void print_Tree()
        {
                for (int i = 0; i < 10; i++) {
                        if (str[i] != null)
                                System.out.print(str[i]);
                        else
                                System.out.print("-");
                }
        }
}

Pseudo Code:

For first case(0—n-1),
if (say)father=p;
then left_son=(2*p)+1;
and right_son=(2*p)+2;
For second case(1—n),
if (say)father=p;
then left_son=(2*p);
and right_son=(2*p)+1;
where father, left_son and right_son are the values of indices of the array.

UML Diagram:

    A(0)    
     /   \
    B(1)  C(2)  
  /   \      \
 D(3)  E(4)   F(6) 
OR,
      A(1)    
     /   \
    B(2)  C(3)  
  /   \      \
 D(4)  E(5)   F(7)

Please Uprate :)