# Check if the given Binary Tree have a Subtree with equal no of 1’s and 0’s

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 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

 // 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

Output:
True

Time Complexity: O(N)
Space Complexity: O(1)

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