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Binary Tree (Array implementation)
  • Difficulty Level : Easy
  • Last Updated : 08 Sep, 2020

Talking about representation, trees can be represented in two ways: 
1) Dynamic Node Representation (Linked Representation). 
2) Array Representation (Sequential Representation).
We are going to talk about the sequential representation of the trees. 
To represent tree using an array, the numbering of nodes can start either from 0–(n-1) or 1– n. 
 

       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)  



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.
 

 

C++




// C++ implementation of tree using array
// numbering starting from 0 to n-1.
#include<bits/stdc++.h>
using namespace std;
char tree[10];
int root(char key)
{
    if(tree[0] != '\0')
        cout << "Tree already had root";
    else
        tree[0] = key;
    return 0;
}
 
int set_left(char key, int parent)
{
    if(tree[parent] == '\0')
        cout << "\nCan't set child at"
             << (parent * 2) + 1
             << " , no parent found";
    else
        tree[(parent * 2) + 1] = key;
    return 0;
}
 
int set_right(char key, int parent)
{
    if(tree[parent] == '\0')
        cout << "\nCan't set child at"
             << (parent * 2) + 2
             << " , no parent found";
    else
        tree[(parent * 2) + 2] = key;
    return 0;
}
 
int print_tree()
{
    cout << "\n";
    for(int i = 0; i < 10; i++)
    {
        if(tree[i] != '\0')
            cout << tree[i];
        else
            cout << "-";
    }
    return 0;
}
 
// Driver Code
int main()
{
    root('A');
    //insert_left('B',0);
    set_right('C', 0);
    set_left('D', 1);
    set_right('E', 1);
    set_right('F', 2);
    print_tree();
    return 0;
}
 
// This code is contributed by
// Gaurav_Kumar_Raghav

Java




// 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("-");
        }
    }
}

C#




// C# implementation of tree using array
// numbering starting from 0 to n-1.
using System;
 
public 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) {
            Console.Write("Can't set child at {0}, 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) {
            Console.Write("Can't set child at {0}, no parent found\n", t);
        }
        else {
            str[t] = key;
        }
    }
 
    public void print_Tree()
    {
        for (int i = 0; i < 10; i++) {
            if (str[i] != null)
                Console.Write(str[i]);
            else
                Console.Write("-");
        }
    }
}
 
// This code contributed by Rajput-Ji

Python3




# Python3 implementation of tree using array
# numbering starting from 0 to n-1.
tree = [None] * 10
 
 
def root(key):
    if tree[0] != None:
        print("Tree already had root")
    else:
        tree[0] = key
 
 
def set_left(key, parent):
    if tree[parent] == None:
        print("Can't set child at", (parent * 2) + 1, ", no parent found")
    else:
        tree[(parent * 2) + 1] = key
 
 
def set_right(key, parent):
    if tree[parent] == None:
        print("Can't set child at", (parent * 2) + 2, ", no parent found")
    else:
        tree[(parent * 2) + 2] = key
 
 
def print_tree():
    for i in range(10):
        if tree[i] != None:
            print(tree[i], end="")
        else:
            print("-", end="")
    print()
 
 
# Driver Code
root('A')
set_right('C', 0)
set_left('D', 1)
set_right('E', 1)
set_right('F', 2)
print_tree()
 
# This code is contributed by Gaurav Kumar Tailor

Output: 
 

Can't set child at 3, no parent found
Can't set child at 4, no parent found
A-C---F---



Note – Please refer this if you want to construct tree from the given parent array.
 

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