Related Articles

# Check if the given binary tree has a sub-tree with equal no of 1’s and 0’s | Set 2

• Difficulty Level : Medium
• Last Updated : 28 Jun, 2021

Given a tree having every node’s value as either 0 or 1, the task is to find whether the given binary tree contains any sub-tree that has equal number of 0’s and 1’s, if such sub-tree is found then print Yes else print No.
Examples:

Input: Output: Yes
There are two sub-trees with equal number of 1’s and 0’s.
Hence the output is “Yes”
Input: Output: No

Approach:

• Update all the nodes of the tree so that they represent the sum of all the nodes in the sub-tree rooted at the current node.
• Now if some node exists whose value is half of the number of nodes in the tree rooted at the same node then it is a valid sub-tree.
• If no such node exists then print No.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// To store whether the tree contains a sub-tree``// with equal number of 0's and 1's``bool` `hasValidSubTree = ``false``;` `// Represents a node of the tree``struct` `node {``    ``int` `data;``    ``struct` `node *right, *left;``};` `// 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 perform inorder traversal on``// the tree and print the nodes in that order``void` `inorder(``struct` `node* root)``{``    ``if` `(root == NULL)``        ``return``;``    ``inorder(root->left);``    ``cout << root->data << endl;``    ``inorder(root->right);``}` `// Function to return the size of the``// sub-tree rooted at the current node``int` `size(``struct` `node* root)``{` `    ``int` `a = 0, b = 0;` `    ``// If root is null or the valid sub-tree``    ``// has already been found``    ``if` `(root == NULL || hasValidSubTree)``        ``return` `0;` `    ``// Size of the right sub-tree``    ``a = size(root->right);` `    ``// 1 is added for the parent``    ``a = a + 1;` `    ``// Size of the left sub-tree``    ``b = size(root->left);` `    ``// Total size of the tree``    ``// rooted at the current node``    ``a = b + a;` `    ``// If the current tree has equal``    ``// number of 0's and 1's``    ``if` `(a % 2 == 0 && root->data == a / 2)``        ``hasValidSubTree = ``true``;` `    ``return` `a;``}` `// Function to update and return the sum``// of all the tree nodes rooted at``// the passed node``int` `sum_tree(``struct` `node* root)``{``    ``if` `(root == NULL)``        ``return` `0;` `    ``int` `a = 0, b = 0;` `    ``// If left child exists``    ``if` `(root->left != NULL)``        ``a = sum_tree(root->left);` `    ``// If right child exists``    ``if` `(root->right != NULL)``        ``b = sum_tree(root->right);``    ``root->data += (a + b);` `    ``return` `root->data;``}` `// Driver code``int` `main()``{``    ``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(1);``    ``root->left->right->right->left->left = newnode(1);` `    ``sum_tree(root);``    ``size(root);` `    ``if` `(hasValidSubTree)``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.Comparator;` `class` `GFG``{`  `// To store whether the tree contains a sub-tree``// with equal number of 0's and 1's``static` `boolean` `hasValidSubTree = ``false``;` `// Represents a node of the tree``static` `class` `node``{``    ``int` `data;``    ``node right, left;``};` `// 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 perform inorder traversal on``// the tree and print the nodes in that order``static` `void` `inorder( node root)``{``    ``if` `(root == ``null``)``        ``return``;``    ``inorder(root.left);``    ``System.out.print(root.data);``    ``inorder(root.right);``}` `// Function to return the size of the``// sub-tree rooted at the current node``static` `int` `size( node root)``{` `    ``int` `a = ``0``, b = ``0``;` `    ``// If root is null or the valid sub-tree``    ``// has already been found``    ``if` `(root == ``null` `|| hasValidSubTree)``        ``return` `0``;` `    ``// Size of the right sub-tree``    ``a = size(root.right);` `    ``// 1 is added for the parent``    ``a = a + ``1``;` `    ``// Size of the left sub-tree``    ``b = size(root.left);` `    ``// Total size of the tree``    ``// rooted at the current node``    ``a = b + a;` `    ``// If the current tree has equal``    ``// number of 0's and 1's``    ``if` `(a % ``2` `== ``0` `&& root.data == a / ``2``)``        ``hasValidSubTree = ``true``;` `    ``return` `a;``}` `// Function to update and return the sum``// of all the tree nodes rooted at``// the passed node``static` `int` `sum_tree( node root)``{``    ``if` `(root == ``null``)``        ``return` `0``;` `    ``int` `a = ``0``, b = ``0``;` `    ``// If left child exists``    ``if` `(root.left != ``null``)``        ``a = sum_tree(root.left);` `    ``// If right child exists``    ``if` `(root.right != ``null``)``        ``b = sum_tree(root.right);``    ``root.data += (a + b);` `    ``return` `root.data;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``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(``1``);``    ``root.left.right.right.left.left = newnode(``1``);` `    ``sum_tree(root);``    ``size(root);` `    ``if` `(hasValidSubTree)``        ``System.out.print( ``"Yes"``);``    ``else``        ``System.out.print( ``"No"``);``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach` `class` `node:``    ` `    ``def` `__init__(``self``, key):``        ``self``.data ``=` `key``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function to perform inorder traversal on``# the tree and print the nodes in that order``def` `inorder(root):` `    ``if` `root ``=``=` `None``:``        ``return``        ` `    ``inorder(root.left)``    ``print``(root.data)``    ``inorder(root.right)` `# Function to return the size of the``# sub-tree rooted at the current node``def` `size(root):` `    ``a, b ``=` `0``, ``0``    ``global` `hasValidSubTree` `    ``# If root is null or the valid``    ``# sub-tree has already been found``    ``if` `root ``=``=` `None` `or` `hasValidSubTree:``        ``return` `0` `    ``# Size of the right sub-tree``    ``a ``=` `size(root.right)` `    ``# 1 is added for the parent``    ``a ``=` `a ``+` `1` `    ``# Size of the left sub-tree``    ``b ``=` `size(root.left)` `    ``# Total size of the tree``    ``# rooted at the current node``    ``a ``=` `b ``+` `a` `    ``# If the current tree has equal``    ``# number of 0's and 1's``    ``if` `a ``%` `2` `=``=` `0` `and` `root.data ``=``=` `a ``/``/` `2``:``        ``hasValidSubTree ``=` `True` `    ``return` `a` `# Function to update and return the sum``# of all the tree nodes rooted at``# the passed node``def` `sum_tree(root):` `    ``if` `root ``=``=` `None``:``        ``return` `0` `    ``a, b ``=` `0``, ``0` `    ``# If left child exists``    ``if` `root.left !``=` `None``:``        ``a ``=` `sum_tree(root.left)` `    ``# If right child exists``    ``if` `root.right !``=` `None``:``        ``b ``=` `sum_tree(root.right)``    ` `    ``root.data ``+``=` `(a ``+` `b)``    ``return` `root.data` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``# To store whether the tree contains a``    ``# sub-tree with equal number of 0's and 1's``    ``hasValidSubTree ``=` `False` `    ``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(``1``)``    ``root.left.right.right.left.left ``=` `node(``1``)` `    ``sum_tree(root)``    ``size(root)` `    ``if` `hasValidSubTree:``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)` `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// To store whether the tree contains a sub-tree``// with equal number of 0's and 1's``static` `bool` `hasValidSubTree = ``false``;` `// Represents a node of the tree``public` `class` `node``{``    ``public` `int` `data;``    ``public` `node right, left;``};` `// 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 perform inorder traversal on``// the tree and print the nodes in that order``static` `void` `inorder( node root)``{``    ``if` `(root == ``null``)``        ``return``;``    ``inorder(root.left);``    ``Console.Write(root.data);``    ``inorder(root.right);``}` `// Function to return the size of the``// sub-tree rooted at the current node``static` `int` `size( node root)``{` `    ``int` `a = 0, b = 0;` `    ``// If root is null or the valid sub-tree``    ``// has already been found``    ``if` `(root == ``null` `|| hasValidSubTree)``        ``return` `0;` `    ``// Size of the right sub-tree``    ``a = size(root.right);` `    ``// 1 is added for the parent``    ``a = a + 1;` `    ``// Size of the left sub-tree``    ``b = size(root.left);` `    ``// Total size of the tree``    ``// rooted at the current node``    ``a = b + a;` `    ``// If the current tree has equal``    ``// number of 0's and 1's``    ``if` `(a % 2 == 0 && root.data == a / 2)``        ``hasValidSubTree = ``true``;` `    ``return` `a;``}` `// Function to update and return the sum``// of all the tree nodes rooted at``// the passed node``static` `int` `sum_tree( node root)``{``    ``if` `(root == ``null``)``        ``return` `0;` `    ``int` `a = 0, b = 0;` `    ``// If left child exists``    ``if` `(root.left != ``null``)``        ``a = sum_tree(root.left);` `    ``// If right child exists``    ``if` `(root.right != ``null``)``        ``b = sum_tree(root.right);``    ``root.data += (a + b);` `    ``return` `root.data;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``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(1);``    ``root.left.right.right.left.left = newnode(1);` `    ``sum_tree(root);``    ``size(root);` `    ``if` `(hasValidSubTree)``        ``Console.Write( ``"Yes"``);``    ``else``        ``Console.Write( ``"No"``);``}``}` `// This code contributed by Rajput-Ji`

## Javascript

 ``
Output:
`No`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up