Given a Binary Tree having data at nodes as either 0's or 1's. The task is to find out whether there exists a subtree having an equal number of 1's and 0's.
Examples:
Input :
Output : True
There are two subtrees present in the above tree where the number of 1's is equal to the number of 0's.Input :
Output : False
There is no such subtree present which has the number of 1's equal to number of 0's
Approach: The idea is to change the data 0's of the tree to -1. So that it becomes very easy to find the subtree having equal number of 0's and 1's. After converting all 0's to -1, create a sum tree. After creating the sum tree, each node will contain the sum of all node lying under it.
Traverse the tree again and find if there is a node having 0 sum, it means that there is a subtree that has the equal number of 1's and -1's, i.e. equal number of 1's and 0's.
Below is the implementation of the above approach:
// C++ program to check if there exist a
// subtree with equal number of 1's and 0's
#include <bits/stdc++.h>
using namespace std;
// Binary Tree Node
struct node {
int data;
struct node *right, *left;
};
// Utility function to create a new node
struct node* newnode(int key)
{
struct node* temp = new node;
temp->data = key;
temp->right = NULL;
temp->left = NULL;
return temp;
}
// Function to convert all 0's in the
// tree to -1
void convert(struct node* root)
{
if (root == NULL) {
return;
}
// Move to right subtree
convert(root->right);
// Replace the 0's with -1 in the tree
if (root->data == 0) {
root->data = -1;
}
// Move to left subtree
convert(root->left);
}
// Function to convert the tree to a SUM tree
int sum_tree(struct node* root)
{
int a = 0, b = 0;
if (root == NULL) {
return 0;
}
a = sum_tree(root->left);
b = sum_tree(root->right);
root->data = root->data + a + b;
return root->data;
}
// Function to check if there exists a subtree
// with equal no of 1s and 0s
int checkSubtree(struct node* root, int d)
{
if (root == NULL) {
return 0;
}
// Check if there is a subtree with equal
// 1s and 0s or not
if (d == 0) {
d = checkSubtree(root->left, d);
}
if (root->data == 0) {
d = 1;
return d;
}
if (d == 0) {
d = checkSubtree(root->right, d);
}
return d;
}
// Driver Code
int main()
{
// Create the Binary Tree
struct node* root = newnode(1);
root->right = newnode(0);
root->right->right = newnode(1);
root->right->right->right = newnode(1);
root->left = newnode(0);
root->left->left = newnode(1);
root->left->left->left = newnode(1);
root->left->right = newnode(0);
root->left->right->left = newnode(1);
root->left->right->left->left = newnode(1);
root->left->right->right = newnode(0);
root->left->right->right->left = newnode(0);
root->left->right->right->left->left = newnode(1);
// Convert all 0s in tree to -1
convert(root);
// Convert the tree into a SUM tree
sum_tree(root);
// Check if required Subtree exists
int d = 0;
if (checkSubtree(root, d)) {
cout << "True" << endl;
}
else {
cout << "False" << endl;
}
return 0;
}
// Java program to check if there exist a
// subtree with equal number of 1's and 0's
import java.util.*;
class GFG{
// Binary Tree Node
static class node {
int data;
node right, left;
};
// Utility function to create a new node
static node newnode(int key)
{
node temp = new node();
temp.data = key;
temp.right = null;
temp.left = null;
return temp;
}
// Function to convert all 0's in the
// tree to -1
static void convert(node root)
{
if (root == null) {
return;
}
// Move to right subtree
convert(root.right);
// Replace the 0's with -1 in the tree
if (root.data == 0) {
root.data = -1;
}
// Move to left subtree
convert(root.left);
}
// Function to convert the tree to a SUM tree
static int sum_tree(node root)
{
int a = 0, b = 0;
if (root == null) {
return 0;
}
a = sum_tree(root.left);
b = sum_tree(root.right);
root.data = root.data + a + b;
return root.data;
}
// Function to check if there exists a subtree
// with equal no of 1s and 0s
static int checkSubtree(node root, int d)
{
if (root == null) {
return 0;
}
// Check if there is a subtree with equal
// 1s and 0s or not
if (d == 0) {
d = checkSubtree(root.left, d);
}
if (root.data == 0) {
d = 1;
return d;
}
if (d == 0) {
d = checkSubtree(root.right, d);
}
return d;
}
// Driver Code
public static void main(String args[])
{
// Create the Binary Tree
node root = newnode(1);
root.right = newnode(0);
root.right.right = newnode(1);
root.right.right.right = newnode(1);
root.left = newnode(0);
root.left.left = newnode(1);
root.left.left.left = newnode(1);
root.left.right = newnode(0);
root.left.right.left = newnode(1);
root.left.right.left.left = newnode(1);
root.left.right.right = newnode(0);
root.left.right.right.left = newnode(0);
root.left.right.right.left.left = newnode(1);
// Convert all 0s in tree to -1
convert(root);
// Convert the tree into a SUM tree
sum_tree(root);
// Check if required Subtree exists
int d = 0;
if (checkSubtree(root, d)>=1) {
System.out.println("True");
}
else {
System.out.println("False");
}
}
}
// This code is contributed by AbhiThakur
# Python3 program to check if there exist a
# subtree with equal number of 1's and 0's
# Binary Tree Node
class node:
def __init__(self, key):
self.data = key
self.left = None
self.right = None
# Function to convert all 0's in the
# tree to -1
def convert(root):
if (root == None):
return
# Move to right subtree
convert(root.right)
# Replace the 0's with -1
# in the tree
if (root.data == 0):
root.data = -1
# Move to left subtree
convert(root.left)
# Function to convert the tree
# to a SUM tree
def sum_tree(root):
a = 0
b = 0
if (root == None):
return 0
a = sum_tree(root.left)
b = sum_tree(root.right)
root.data = root.data + a + b
return root.data
# Function to check if there exists
# a subtree with equal no of 1s and 0s
def checkSubtree(root, d):
if (root == None):
return 0
# Check if there is a subtree with
# equal 1s and 0s or not
if (d == 0):
d = checkSubtree(root.left, d)
if (root.data == 0):
d = 1
return d
if (d == 0):
d = checkSubtree(root.right, d)
return d
# Driver Code
if __name__ == '__main__':
# Create the Binary Tree
root = node(1)
root.right = node(0)
root.right.right = node(1)
root.right.right.right = node(1)
root.left = node(0)
root.left.left = node(1)
root.left.left.left = node(1)
root.left.right = node(0)
root.left.right.left = node(1)
root.left.right.left.left = node(1)
root.left.right.right = node(0)
root.left.right.right.left = node(0)
root.left.right.right.left.left = node(1)
# Convert all 0s in tree to -1
convert(root)
# Convert the tree into a SUM tree
sum_tree(root)
# Check if required Subtree exists
d = 0
if (checkSubtree(root, d)):
print("True")
else:
print("False")
# This code is contributed by mohit kumar 29
// C# program to check if there exist a
// subtree with equal number of 1's and 0's
using System;
class GFG{
// Binary Tree Node
class node {
public int data;
public node right, left;
};
// Utility function to create a new node
static node newnode(int key)
{
node temp = new node();
temp.data = key;
temp.right = null;
temp.left = null;
return temp;
}
// Function to convert all 0's in the
// tree to -1
static void convert(node root)
{
if (root == null) {
return;
}
// Move to right subtree
convert(root.right);
// Replace the 0's with -1 in the tree
if (root.data == 0) {
root.data = -1;
}
// Move to left subtree
convert(root.left);
}
// Function to convert the tree to a SUM tree
static int sum_tree(node root)
{
int a = 0, b = 0;
if (root == null) {
return 0;
}
a = sum_tree(root.left);
b = sum_tree(root.right);
root.data = root.data + a + b;
return root.data;
}
// Function to check if there exists a subtree
// with equal no of 1s and 0s
static int checkSubtree(node root, int d)
{
if (root == null) {
return 0;
}
// Check if there is a subtree with equal
// 1s and 0s or not
if (d == 0) {
d = checkSubtree(root.left, d);
}
if (root.data == 0) {
d = 1;
return d;
}
if (d == 0) {
d = checkSubtree(root.right, d);
}
return d;
}
// Driver Code
public static void Main(String []args)
{
// Create the Binary Tree
node root = newnode(1);
root.right = newnode(0);
root.right.right = newnode(1);
root.right.right.right = newnode(1);
root.left = newnode(0);
root.left.left = newnode(1);
root.left.left.left = newnode(1);
root.left.right = newnode(0);
root.left.right.left = newnode(1);
root.left.right.left.left = newnode(1);
root.left.right.right = newnode(0);
root.left.right.right.left = newnode(0);
root.left.right.right.left.left = newnode(1);
// Convert all 0s in tree to -1
convert(root);
// Convert the tree into a SUM tree
sum_tree(root);
// Check if required Subtree exists
int d = 0;
if (checkSubtree(root, d) >= 1) {
Console.WriteLine("True");
}
else {
Console.WriteLine("False");
}
}
}
// This code is contributed by sapnasingh4991
<script>
// Javascript program to check if there exist a
// subtree with equal number of 1's and 0's
class Node
{
constructor(key)
{
this.data = key;
this.left = null;
this.right = null;
}
}
// Function to convert all 0's in the
// tree to -1
function convert(root)
{
if (root == null)
{
return;
}
// Move to right subtree
convert(root.right);
// Replace the 0's with -1 in the tree
if (root.data == 0)
{
root.data = -1;
}
// Move to left subtree
convert(root.left);
}
// Function to convert the tree to a SUM tree
function sum_tree(root)
{
let a = 0, b = 0;
if (root == null)
{
return 0;
}
a = sum_tree(root.left);
b = sum_tree(root.right);
root.data = root.data + a + b;
return root.data;
}
// Function to check if there exists a subtree
// with equal no of 1s and 0s
function checkSubtree(root, d)
{
if (root == null)
{
return 0;
}
// Check if there is a subtree with equal
// 1s and 0s or not
if (d == 0)
{
d = checkSubtree(root.left, d);
}
if (root.data == 0)
{
d = 1;
return d;
}
if (d == 0)
{
d = checkSubtree(root.right, d);
}
return d;
}
// Driver Code
// Create the Binary Tree
let root = new Node(1);
root.right = new Node(0);
root.right.right = new Node(1);
root.right.right.right = new Node(1);
root.left = new Node(0);
root.left.left = new Node(1);
root.left.left.left = new Node(1);
root.left.right = new Node(0);
root.left.right.left = new Node(1);
root.left.right.left.left = new Node(1);
root.left.right.right = new Node(0);
root.left.right.right.left = new Node(0);
root.left.right.right.left.left = new Node(1);
// Convert all 0s in tree to -1
convert(root);
// Convert the tree into a SUM tree
sum_tree(root);
// Check if required Subtree exists
let d = 0;
if (checkSubtree(root, d) >= 1)
{
document.write("True<br>");
}
else
{
document.write("False<br>");
}
// This code is contributed by unknown2108
</script>
Output:
True
Time Complexity: O(N)
Space Complexity: O(1)