Binary Tree (Array implementation)

Last Updated : 06 Apr, 2023

Given an array that represents a tree in such a way that array indexes are values in tree nodes and array values give the parent node of that particular index (or node). The value of the root node index would always be -1 as there is no parent for root. Construct the standard linked representation of given Binary Tree from this given representation. Do refer in order to understand how to construct binary tree from given parent array representation.

Ways to represent:

Trees can be represented in two ways as listed below:

Dynamic Node Representation (Linked Representation).
Array Representation (Sequential Representation).

Now, we are going to talk about the sequential representation of the trees. In order to represent a tree using an array, the numbering of nodes can start either from 0--(n-1) or 1-- n, consider the below illustration as follows:

Illustration:

       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)

Procedure:

Note: father, left_son and right_son are the values of indices of the array.

Case 1: (0---n-1)

if (say)father=p;  then left_son=(2*p)+1;  and right_son=(2*p)+2;

Case 2: 1---n

if (say)father=p;  then left_son=(2*p);  and right_son=(2*p)+1;

Implementation:

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');   set_left('B',0);   set_right('C', 0);   set_left('D', 1);   set_right('E', 1);   set_right('F', 2);   print_tree();   return 0; }

Java

// JAVA implementation of tree using array // numbering starting from 0 to n-1.  // Importing required classes import java.io.*; import java.lang.*; import java.util.*;  // Class 1 // Helper class (Node class) public class Tree {      // Main driver method     public static void main(String[] args)     {          // Creating object of class 2 inside main() method         Array_imp obj = new Array_imp();          // Setting root node         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_Right("F", 2);         obj.print_Tree();     } }  // Class 2 // Helper class class Array_imp {      // Member variables of this class     static int root = 0;     static String[] str = new String[10];      // Method 1     // Creating root node     public void Root(String key) { str[0] = key; }      // Method 2     // Creating 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;         }     }      // Method 3     // Creating 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;         }     }      // Method 4     // To print our tree     public void print_Tree()     {          // Iterating using for loop         for (int i = 0; i < 10; i++) {             if (str[i] != null)                 System.out.print(str[i]);             else                 System.out.print("-");         }     } }

// 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_Right("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_left('B', 0) 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

JavaScript

// JavaScript implementation of tree using array // numbering starting from 0 to n-1. const tree = Array(10).fill(null);  function root(key) {     if (tree[0] != null) {         console.log("Tree already had root");     } else {         tree[0] = key;     } }  function setLeft(key, parent) {     if (tree[parent] == null) {         console.log(`Can't set child at ${(parent * 2) + 1}, no parent found <br>`);     } else {         tree[(parent * 2) + 1] = key;     } }  function setRight(key, parent) {     if (tree[parent] == null) {         console.log(`Can't set child at ${(parent * 2) + 2}, no parent found <br>`);     } else {         tree[(parent * 2) + 2] = key;     } }  function printTree() {     for (let i = 0; i < 10; i++) {         if (tree[i] != null) {             console.log(tree[i]);         } else {             console.log("-");         }     } }  // Driver Code root("A"); setLeft("B", 0); setRight("C", 0); setLeft("D", 1); setRight("E", 1); setRight("F", 2); printTree();    // This code is contributed by lokeshpotta20

Output

ABCDE-F---

Time complexity: O(log n) since using heap to create a binary tree
Space complexity: O(1)

Maximum Depth of Binary Tree

sanjal_katiyar

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