Count subtrees that sum up to a given value x

Given a binary tree containing n nodes. The problem is to count subtress having total node’s data sum equal to a given value x.

Examples:

Input : 
             5
           /   \  
        -10     3
        /  \   /  \
       9    8 -4   7
       
       x = 7

Output : 2
There are 2 subtrees with sum 7.

1st one is leaf node:
7.

2nd one is:

      -10
     /   \
    9     8

Source: Microsoft Interview Experience | Set 157.

Recommended: Please solve it on PRACTICE first, before moving on to the solution.

Algorithm:

countSubtreesWithSumX(root, count, x)
    if !root then
        return 0
        
    ls = countSubtreesWithSumX(root->left, count, x)
    rs = countSubtreesWithSumX(root->right, count, x)
    sum = ls + rs + root->data
    
    if sum == x then
    count++
    return sum

countSubtreesWithSumXUtil(root, x)
    if !root then
        return 0
    
    Initialize count = 0
    ls = countSubtreesWithSumX(root->left, count, x)
    rs = countSubtreesWithSumX(root->right, count, x)
    
    if (ls + rs + root->data) == x
        count++
    return count

C++

// C++ implementation to count subtress that 
// sum up to a given value x 
#include <bits/stdc++.h> 
  
using namespace std; 
  
// structure of a node of binary tree 
struct Node { 
    int data; 
    Node *left, *right; 
}; 
  
// function to get a new node 
Node* getNode(int data) 
{ 
    // allocate space 
    Node* newNode = (Node*)malloc(sizeof(Node)); 
  
    // put in the data 
    newNode->data = data; 
    newNode->left = newNode->right = NULL; 
    return newNode; 
} 
  
// function to count subtress that 
// sum up to a given value x 
int countSubtreesWithSumX(Node* root, 
                          int& count, int x) 
{ 
    // if tree is empty 
    if (!root) 
        return 0; 
  
    // sum of nodes in the left subtree 
    int ls = countSubtreesWithSumX(root->left, count, x); 
  
    // sum of nodes in the right subtree 
    int rs = countSubtreesWithSumX(root->right, count, x); 
  
    // sum of nodes in the subtree rooted 
    // with 'root->data' 
    int sum = ls + rs + root->data; 
  
    // if true 
    if (sum == x) 
        count++; 
  
    // return subtree's nodes sum 
    return sum; 
} 
  
// utility function to count subtress that 
// sum up to a given value x 
int countSubtreesWithSumXUtil(Node* root, int x) 
{ 
    // if tree is empty 
    if (!root) 
        return 0; 
  
    int count = 0; 
  
    // sum of nodes in the left subtree 
    int ls = countSubtreesWithSumX(root->left, count, x); 
  
    // sum of nodes in the right subtree 
    int rs = countSubtreesWithSumX(root->right, count, x); 
  
    // if tree's nodes sum == x 
    if ((ls + rs + root->data) == x) 
        count++; 
  
    // required count of subtrees 
    return count; 
} 
  
// Driver program to test above 
int main() 
{ 
    /* binary tree creation     
                5 
              /   \   
           -10     3 
           /  \   /  \ 
          9    8 -4   7 
    */
    Node* root = getNode(5); 
    root->left = getNode(-10); 
    root->right = getNode(3); 
    root->left->left = getNode(9); 
    root->left->right = getNode(8); 
    root->right->left = getNode(-4); 
    root->right->right = getNode(7); 
  
    int x = 7; 
  
    cout << "Count = "
         << countSubtreesWithSumXUtil(root, x); 
  
    return 0; 
} 

Java

// Java program to find if  
// there is a subtree with  
// given sum  
import java.util.*;  
class GFG 
{ 
  
// structure of a node  
// of binary tree  
static class Node 
{  
    int data;  
    Node left, right;  
} 
  
static class INT 
{ 
    int v; 
    INT(int a) 
    { 
        v = a; 
    } 
} 
  
// function to get a new node  
static Node getNode(int data)  
{  
    // allocate space  
    Node newNode = new Node();  
  
    // put in the data  
    newNode.data = data;  
    newNode.left = newNode.right = null;  
    return newNode;  
}  
  
// function to count subtress that  
// sum up to a given value x  
static int countSubtreesWithSumX(Node root,  
                          INT count, int x)  
{  
    // if tree is empty  
    if (root == null)  
        return 0;  
  
    // sum of nodes in the left subtree  
    int ls = countSubtreesWithSumX(root.left,  
                                   count, x);  
  
    // sum of nodes in the right subtree  
    int rs = countSubtreesWithSumX(root.right,  
                                   count, x);  
  
    // sum of nodes in the subtree  
    // rooted with 'root.data'  
    int sum = ls + rs + root.data;  
  
    // if true  
    if (sum == x)  
        count.v++;  
  
    // return subtree's nodes sum  
    return sum;  
}  
  
// utility function to  
// count subtress that  
// sum up to a given value x  
static int countSubtreesWithSumXUtil(Node root,  
                                     int x)  
{  
    // if tree is empty  
    if (root == null)  
        return 0;  
  
    INT count = new INT(0);  
  
    // sum of nodes in the left subtree  
    int ls = countSubtreesWithSumX(root.left, 
                                   count, x);  
  
    // sum of nodes in the right subtree  
    int rs = countSubtreesWithSumX(root.right,  
                                   count, x);  
  
    // if tree's nodes sum == x  
    if ((ls + rs + root.data) == x)  
        count.v++;  
  
    // required count of subtrees  
    return count.v;  
}  
  
// Driver Code 
public static void main(String args[]) 
{  
    /* binary tree creation      
                5  
            / \  
        -10     3  
        / \ / \  
        9 8 -4 7  
    */
    Node root = getNode(5);  
    root.left = getNode(-10);  
    root.right = getNode(3);  
    root.left.left = getNode(9);  
    root.left.right = getNode(8);  
    root.right.left = getNode(-4);  
    root.right.right = getNode(7);  
  
    int x = 7;  
  
    System.out.println("Count = " +  
           countSubtreesWithSumXUtil(root, x));  
}  
} 
  
// This code is contributed  
// by Arnab Kundu 

Python3

# Python3 implementation to count subtress
# that Sum up to a given value x

# class to get a new node
class getNode:
def __init__(self, data):

# put in the data
self.data = data
self.left = self.right = None

# function to count subtress that
# Sum up to a given value x
def countSubtreesWithSumX(root, count, x):

# if tree is empty
if (not root):
return 0

# Sum of nodes in the left subtree
ls = countSubtreesWithSumX(root.left,
count, x)

# Sum of nodes in the right subtree
rs = countSubtreesWithSumX(root.right,
count, x)

# Sum of nodes in the subtree
# rooted with ‘root.data’
Sum = ls + rs + root.data

# if true
if (Sum == x):
count[0] += 1

# return subtree’s nodes Sum
return Sum

# utility function to count subtress
# that Sum up to a given value x
def countSubtreesWithSumXUtil(root, x):

# if tree is empty
if (not root):
return 0

count = [0]

# Sum of nodes in the left subtree
ls = countSubtreesWithSumX(root.left,
count, x)

# Sum of nodes in the right subtree
rs = countSubtreesWithSumX(root.right,
count, x)

# if tree’s nodes Sum == x
if ((ls + rs + root.data) == x):
count[0] += 1

# required count of subtrees
return count[0]

# Driver Code
if __name__ == ‘__main__’:

# binary tree creation
# 5
# / \
# -10 3
# / \ / \
# 9 8 -4 7
root = getNode(5)
root.left = getNode(-10)
root.right = getNode(3)
root.left.left = getNode(9)
root.left.right = getNode(8)
root.right.left = getNode(-4)
root.right.right = getNode(7)

x = 7

print(“Count =”,
countSubtreesWithSumXUtil(root, x))

# This code is contributed by PranchalK

C#

using System; 
  
// c# program to find if  
// there is a subtree with  
// given sum  
public class GFG 
{ 
  
// structure of a node  
// of binary tree  
public class Node 
{ 
    public int data; 
    public Node left, right; 
} 
  
public class INT 
{ 
    public int v; 
    public INT(int a) 
    { 
        v = a; 
    } 
} 
  
// function to get a new node  
public static Node getNode(int data) 
{ 
    // allocate space  
    Node newNode = new Node(); 
  
    // put in the data  
    newNode.data = data; 
    newNode.left = newNode.right = null; 
    return newNode; 
} 
  
// function to count subtress that  
// sum up to a given value x  
public static int countSubtreesWithSumX(Node root,  
                                    INT count, int x) 
{ 
    // if tree is empty  
    if (root == null) 
    { 
        return 0; 
    } 
  
    // sum of nodes in the left subtree  
    int ls = countSubtreesWithSumX(root.left, count, x); 
  
    // sum of nodes in the right subtree  
    int rs = countSubtreesWithSumX(root.right, count, x); 
  
    // sum of nodes in the subtree  
    // rooted with 'root.data'  
    int sum = ls + rs + root.data; 
  
    // if true  
    if (sum == x) 
    { 
        count.v++; 
    } 
  
    // return subtree's nodes sum  
    return sum; 
} 
  
// utility function to  
// count subtress that  
// sum up to a given value x  
public static int countSubtreesWithSumXUtil(Node root, int x) 
{ 
    // if tree is empty  
    if (root == null) 
    { 
        return 0; 
    } 
  
    INT count = new INT(0); 
  
    // sum of nodes in the left subtree  
    int ls = countSubtreesWithSumX(root.left, count, x); 
  
    // sum of nodes in the right subtree  
    int rs = countSubtreesWithSumX(root.right, count, x); 
  
    // if tree's nodes sum == x  
    if ((ls + rs + root.data) == x) 
    { 
        count.v++; 
    } 
  
    // required count of subtrees  
    return count.v; 
} 
  
// Driver Code  
public static void Main(string[] args) 
{ 
    /* binary tree creation      
                5  
            / \  
        -10     3  
        / \ / \  
        9 8 -4 7  
    */
    Node root = getNode(5); 
    root.left = getNode(-10); 
    root.right = getNode(3); 
    root.left.left = getNode(9); 
    root.left.right = getNode(8); 
    root.right.left = getNode(-4); 
    root.right.right = getNode(7); 
  
    int x = 7; 
  
    Console.WriteLine("Count = " + countSubtreesWithSumXUtil(root, x)); 
} 
} 
  
// This code is contributed by Shrikant13 

Output:

Count = 2

Time Complexity: O(n).

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Recommended: Please solve it on PRACTICE first, before moving on to the solution.

C++

Java

Python3

C#

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