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++
// 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; } |
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Java
// 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 |
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C#
// 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 |
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Output:
True
Time Complexity: O(N)
Space Complexity: O(1)
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