Number of subtrees having odd count of even numbers

Given a binary tree, find the number of subtrees having odd count of even numbers.

Examples:

Input :         2             
             /     \          
           1        3            
         /   \     /  \       
        4    10   8     5     
             /                
            6      
Output : 6
The subtrees are 4, 6,    1,      8,    3
                        /   \          /  \
                        4    10       8    5
                            /
                           6

               2             
             /     \          
           1        3            
         /   \     /  \       
        4    10   8     5     
             /                
            6   

The idea is to recursively traverse the tree. For every node, recursively count even numbers in left and right subtrees. If root is also even, then add it also to count. If count becomes odd, increment result. 

count = 0 // Initialize result

int countSubtrees(root)
{
  if (root == NULL)
    return 0;

  // count even numbers in left subtree
  int c = countSubtrees(root-> left);

  // Add count of even numbers in right subtree
  c += countSubtrees(root-> right);

  // check if root data is an even number
  if (root-> data % 2 == 0)
    c += 1;

  // If total count of even numbers
  // for the subtree is odd
  if (c % 2 != 0)
    count++;

  // return total count of even
  // numbers of the subtree
  return c;
}

Implementation:

C++

// C++ implementation to find number of
// subtrees having odd count of even numbers
#include <bits/stdc++.h>
using namespace std;

/* A binary tree Node */
struct Node
{
    int data;
    struct Node* left, *right;
};

/* Helper function that allocates a new Node with the
   given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return(node);
}

// Returns count of subtrees having odd count of
// even numbers
int countRec(struct Node* root, int *pcount)
{
    // base condition
    if (root == NULL)
        return 0;

    // count even nodes in left subtree
    int c = countRec(root->left, pcount);

    // Add even nodes in right subtree
    c += countRec(root->right, pcount);

    // Check if root data is an even number
    if (root->data % 2 == 0)
        c += 1;

    // if total count of even numbers
    // for the subtree is odd
    if (c % 2 != 0)
        (*pcount)++;

    // Total count of even nodes of the subtree
    return c;
}

// A wrapper over countRec()
int countSubtrees(Node *root)
{
    int count = 0;
    int *pcount = &count;
    countRec(root, pcount);
    return count;
}

// Driver program to test above
int main()
{
    // binary tree formation
    struct Node *root = newNode(2);   /*          2         */
    root->left        = newNode(1);   /*      /     \       */
    root->right       = newNode(3);   /*     1        3        */
    root->left->left  = newNode(4);   /*   /   \     /  \    */
    root->left->right = newNode(10);  /*  4    10   8     5  */
    root->right->left  = newNode(8);  /*       /             */
    root->right->right = newNode(5);  /*     6               */
    root->left->right->left = newNode(6);

    cout<<"Count = "<< countSubtrees(root);
    return 0;
}

// This code is contributed by famously.

C

// C implementation to find number of
// subtrees having odd count of even numbers
#include <bits/stdc++.h>

/* A binary tree Node */
struct Node
{
    int data;
    struct Node* left, *right;
};

/* Helper function that allocates a new Node with the
   given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return(node);
}

// Returns count of subtrees having odd count of
// even numbers
int countRec(struct Node* root, int *pcount)
{
    // base condition
    if (root == NULL)
        return 0;

    // count even nodes in left subtree
    int c = countRec(root->left, pcount);

    // Add even nodes in right subtree
    c += countRec(root->right, pcount);

    // Check if root data is an even number
    if (root->data % 2 == 0)
        c += 1;

    // if total count of even numbers
    // for the subtree is odd
    if (c % 2 != 0)
        (*pcount)++;

    // Total count of even nodes of the subtree
    return c;
}

// A wrapper over countRec()
int countSubtrees(Node *root)
{
    int count = 0;
    int *pcount = &count;
    countRec(root, pcount);
    return count;
}

// Driver program to test above
int main()
{
    // binary tree formation
    struct Node *root = newNode(2);   /*          2         */
    root->left        = newNode(1);   /*      /     \       */
    root->right       = newNode(3);   /*     1        3        */
    root->left->left  = newNode(4);   /*   /   \     /  \    */
    root->left->right = newNode(10);  /*  4    10   8     5  */
    root->right->left  = newNode(8);  /*       /             */
    root->right->right = newNode(5);  /*     6               */
    root->left->right->left = newNode(6);

    printf("Count = %d", countSubtrees(root));
    return 0;
}

Java

// Java implementation to find number of
// subtrees having odd count of even numbers
import java.util.*;
class GfG {

/* A binary tree Node */
static class Node
{
    int data;
    Node left, right;
}

/* Helper function that allocates a new Node with the
given data and NULL left and right pointers. */
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;
    return(node);
}

// Returns count of subtrees having odd count of
// even numbers
static class P
{
    int pcount = 0;
}
static int countRec(Node root, P p)
{
    // base condition
    if (root == null)
        return 0;

    // count even nodes in left subtree
    int c = countRec(root.left, p);

    // Add even nodes in right subtree
    c += countRec(root.right, p);

    // Check if root data is an even number
    if (root.data % 2 == 0)
        c += 1;

    // if total count of even numbers
    // for the subtree is odd
    if (c % 2 != 0)
        (p.pcount)++;

    // Total count of even nodes of the subtree
    return c;
}

// A wrapper over countRec()
static int countSubtrees(Node root)
{
    P p = new P();
    countRec(root, p);
    return p.pcount;
}

// Driver program to test above
public static void main(String[] args)
{
    // binary tree formation
     Node root = newNode(2); /*         2         */
    root.left     = newNode(1); /*     /     \     */
    root.right     = newNode(3); /*     1     3     */
    root.left.left = newNode(4); /* / \     / \ */
    root.left.right = newNode(10); /* 4 10 8     5 */
    root.right.left = newNode(8); /*     /             */
    root.right.right = newNode(5); /*     6             */
    root.left.right.left = newNode(6);

System.out.println("Count = " + countSubtrees(root));
}
}

Python3

# Python3 implementation to find number of
# subtrees having odd count of even numbers

# Helper class that allocates a new Node
# with the given data and None left and
# right pointers.
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None

# Returns count of subtrees having odd
# count of even numbers
def countRec(root, pcount):

    # base condition
    if (root == None):
        return 0

    # count even nodes in left subtree
    c = countRec(root.left, pcount)

    # Add even nodes in right subtree
    c += countRec(root.right, pcount)

    # Check if root data is an even number
    if (root.data % 2 == 0):
        c += 1

    # if total count of even numbers
    # for the subtree is odd
    if c % 2 != 0:
        pcount[0] += 1

    # Total count of even nodes of
    # the subtree
    return c

# A wrapper over countRec()
def countSubtrees(root):
    count = [0]
    pcount = count
    countRec(root, pcount)
    return count[0]

# Driver Code
if __name__ == '__main__':

    # binary tree formation
    root = newNode(2) #         2
    root.left     = newNode(1) #     /     \
    root.right     = newNode(3) #     1     3
    root.left.left = newNode(4) # / \     / \
    root.left.right = newNode(10) # 4 10 8     5
    root.right.left = newNode(8) #     /
    root.right.right = newNode(5) #     6
    root.left.right.left = newNode(6)

    print("Count = ", countSubtrees(root))

# This code is contributed by PranchalK

C#

// C# implementation to find number of
// subtrees having odd count of even numbers
using System;

class GFG
{

/* A binary tree Node */
public class Node
{
    public int data;
    public Node left, right;
}

/* Helper function that allocates
a new Node with the given data and
NULL left and right pointers. */
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;
    return(node);
}

// Returns count of subtrees
// having odd count of even numbers
public class P
{
    public int pcount = 0;
}

static int countRec(Node root, P p)
{
    // base condition
    if (root == null)
        return 0;

    // count even nodes in left subtree
    int c = countRec(root.left, p);

    // Add even nodes in right subtree
    c += countRec(root.right, p);

    // Check if root data is an even number
    if (root.data % 2 == 0)
        c += 1;

    // if total count of even numbers
    // for the subtree is odd
    if (c % 2 != 0)
        (p.pcount)++;

    // Total count of even nodes of the subtree
    return c;
}

// A wrapper over countRec()
static int countSubtrees(Node root)
{
    P p = new P();
    countRec(root, p);
    return p.pcount;
}

// Driver Code
public static void Main(String[] args)
{
    // binary tree formation
    Node root = newNode(2); /*         2         */
    root.left     = newNode(1); /*     /     \     */
    root.right     = newNode(3); /*     1     3     */
    root.left.left = newNode(4); /* / \     / \ */
    root.left.right = newNode(10); /* 4 10 8     5 */
    root.right.left = newNode(8); /*     /             */
    root.right.right = newNode(5); /*     6             */
    root.left.right.left = newNode(6);

    Console.WriteLine("Count = " +
                       countSubtrees(root));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation to find number of
// subtrees having odd count of even numbers

// A binary tree Node
class Node
{

    // Helper function that allocates a new
    // Node with the given data and NULL left
    // and right pointers.
    constructor(data)
    {
        this.data = data;
        this.left = this.right = null;
    }
}

// Returns count of subtrees having odd
// count of even numbers
class P
{
    constructor()
    {
        this.pcount = 0;
    }
}

function countRec(root, p)
{

    // Base condition
    if (root == null)
        return 0;

    // Count even nodes in left subtree
    let c = countRec(root.left, p);

    // Add even nodes in right subtree
    c += countRec(root.right, p);

    // Check if root data is an even number
    if (root.data % 2 == 0)
        c += 1;

    // If total count of even numbers
    // for the subtree is odd
    if (c % 2 != 0)
        (p.pcount)++;

    // Total count of even nodes
    // of the subtree
    return c;
}

// A wrapper over countRec()
function countSubtrees(root)
{
    let p = new P();
    countRec(root, p);
    return p.pcount;
}

// Driver code

// Binary tree formation
let root = new Node(2); /*               2         */
root.left = new Node(1); /*           /     \      */
root.right = new Node(3); /*         1       3     */
root.left.left = new Node(4); /*    / \     / \    */
root.left.right = new Node(10); /* 4   10  8   5   */
root.right.left = new Node(8); /*     /            */
root.right.right = new Node(5); /*   6             */
root.left.right.left = new Node(6);

document.write("Count = " + countSubtrees(root) + "<br>");

// This code is contributed by rag2127

</script>

Output

Count = 6

Time complexity: O(n) // where n is the number of nodes in the binary tree.

Space complexity: O(n)