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

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 <iostream>
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# 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

<script>

    // JavaScript implementation of the approach
    
    // To store whether the tree contains a sub-tree 
    // with equal number of 0's and 1's 
    let hasValidSubTree = false; 
    
    // Binary tree node
    class node
    {
        constructor(key) {
           this.left = null;
           this.right = null;
           this.data = key;
        }
    }
    
    // To create a new node 
    function newnode(key) 
    { 
        let temp = new node(key);
        return temp; 
    } 

    // Function to perform inorder traversal on 
    // the tree and print the nodes in that order 
    function inorder(root) 
    { 
        if (root == null) 
            return; 
        inorder(root.left); 
        document.write(root.data); 
        inorder(root.right); 
    } 

    // Function to return the size of the 
    // sub-tree rooted at the current node 
    function size(root) 
    { 

        let 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 == parseInt(a / 2, 10)) 
            hasValidSubTree = true; 

        return a; 
    } 

    // Function to update and return the sum 
    // of all the tree nodes rooted at 
    // the passed node 
    function sum_tree(root) 
    { 
        if (root == null) 
            return 0; 

        let 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; 
    } 
    
    let 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) 
        document.write( "Yes"); 
    else
        document.write( "No"); 

</script>

Output:

No

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

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

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