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++
#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;
}
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
import java.io.*;
import java.lang.*;
import java.util.*;
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];
public void Root(String key) { str[0] = key; }
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;
}
}
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#
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];
public void Root(String key)
{
str[0] = key;
}
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;
}
}
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("-");
}
}
}
|
Python3
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()
root('A')
set_left('B', 0)
set_right('C', 0)
set_left('D', 1)
set_right('E', 1)
set_right('F', 2)
print_tree()
|
Javascript
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("-");
}
}
}
root("A");
setLeft("B", 0);
setRight("C", 0);
setLeft("D", 1);
setRight("E", 1);
setRight("F", 2);
printTree();
|
Time complexity: O(log n) since using heap to create a binary tree
Space complexity: O(1)
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