Subtree with given sum in a Binary Tree

You are given a binary tree and a given sum. The task is to check if there exist a subtree whose sum of all nodes is equal to the given sum.

Examples :

// For above tree
Input : sum = 17
Output: "Yes"
// sum of all nodes of subtree {3, 5, 9} = 17

Input : sum = 11
Output: "No"
// no subtree with given sum exist

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to traverse tree in Postorder fashion because here we have to think bottom-up . First calculate the sum of left subtree then right subtree and check if sum_left + sum_right + cur_node = sum is satisfying the condition that means any subtree with given sum exist. Below is the recursive implementation of algorithm.

C++

// C++ program to find if there is a subtree with 
// given sum 
#include<bits/stdc++.h> 
using namespace std; 
  
/* A binary tree node has data, pointer to left child 
   and a pointer to right child */
struct Node 
{ 
    int data; 
    struct Node* left, *right; 
}; 
  
/* utility 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); 
} 
  
// function to check if there exist any subtree with given sum 
// cur_sum  --> sum of current subtree from ptr as root 
// sum_left --> sum of left subtree from ptr as root 
// sum_right --> sum of right subtree from ptr as root 
bool sumSubtreeUtil(struct Node *ptr, int *cur_sum, int sum) 
{ 
    // base condition 
    if (ptr == NULL) 
    { 
        *cur_sum = 0; 
        return false; 
    } 
  
    // Here first we go to left sub-tree, then right subtree 
    // then first we calculate sum of all nodes of subtree 
    // having ptr as root and assign it as cur_sum 
    // cur_sum = sum_left + sum_right + ptr->data 
    // after that we check if cur_sum == sum 
    int sum_left = 0, sum_right = 0; 
    return ( sumSubtreeUtil(ptr->left, &sum_left, sum) || 
             sumSubtreeUtil(ptr->right, &sum_right, sum) || 
        ((*cur_sum = sum_left + sum_right + ptr->data) == sum)); 
} 
  
// Wrapper over sumSubtreeUtil() 
bool sumSubtree(struct Node *root, int sum) 
{ 
    // Initialize sum of subtree with root 
    int cur_sum = 0; 
  
    return sumSubtreeUtil(root, &cur_sum, sum); 
} 
  
// driver program to run the case 
int main() 
{ 
    struct Node *root = newnode(8); 
    root->left    = newnode(5); 
    root->right   = newnode(4); 
    root->left->left = newnode(9); 
    root->left->right = newnode(7); 
    root->left->right->left = newnode(1); 
    root->left->right->right = newnode(12); 
    root->left->right->right->right = newnode(2); 
    root->right->right = newnode(11); 
    root->right->right->left = newnode(3); 
    int sum = 22; 
  
    if (sumSubtree(root, sum)) 
        cout << "Yes"; 
    else
        cout << "No"; 
    return 0; 
} 

Java

// Java program to find if there  
// is a subtree with given sum  
import java.util.*;  
class GFG 
{ 
  
/* A binary tree node has data,  
pointer to left child and a 
pointer to right child */
static class Node  
{  
    int data;  
    Node left, right;  
} 
  
static class INT 
{ 
    int v; 
    INT(int a) 
    { 
        v = a; 
    } 
} 
  
/* utility 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 = node.right = null;  
    return (node);  
}  
  
// function to check if there exist  
// any subtree with given sum  
// cur_sum -. sum of current subtree  
//            from ptr as root  
// sum_left -. sum of left subtree 
//             from ptr as root  
// sum_right -. sum of right subtree 
//              from ptr as root  
static boolean sumSubtreeUtil(Node ptr,  
                              INT cur_sum,  
                              int sum)  
{  
    // base condition  
    if (ptr == null)  
    {  
        cur_sum = new INT(0);  
        return false;  
    }  
  
    // Here first we go to left  
    // sub-tree, then right subtree  
    // then first we calculate sum  
    // of all nodes of subtree having  
    // ptr as root and assign it as  
    // cur_sum. (cur_sum = sum_left +  
    // sum_right + ptr.data) after that 
    // we check if cur_sum == sum  
    INT sum_left = new INT(0),  
        sum_right = new INT(0);  
    return (sumSubtreeUtil(ptr.left, sum_left, sum) ||  
            sumSubtreeUtil(ptr.right, sum_right, sum) ||  
        ((cur_sum.v = sum_left.v +  
                      sum_right.v + ptr.data) == sum));  
}  
  
// Wrapper over sumSubtreeUtil()  
static boolean sumSubtree(Node root, int sum)  
{  
    // Initialize sum of  
    // subtree with root  
    INT cur_sum = new INT( 0);  
  
    return sumSubtreeUtil(root, cur_sum, sum);  
}  
  
// Driver Code 
public static void main(String args[]) 
{  
    Node root = newnode(8);  
    root.left = newnode(5);  
    root.right = newnode(4);  
    root.left.left = newnode(9);  
    root.left.right = newnode(7);  
    root.left.right.left = newnode(1);  
    root.left.right.right = newnode(12);  
    root.left.right.right.right = newnode(2);  
    root.right.right = newnode(11);  
    root.right.right.left = newnode(3);  
    int sum = 22;  
  
    if (sumSubtree(root, sum))  
        System.out.println( "Yes");  
    else
        System.out.println( "No");  
}  
} 
  
// This code is contributed  
// by Arnab Kundu 

C#

using System; 
  
// C# program to find if there  
// is a subtree with given sum  
public class GFG 
{ 
  
/* A binary tree node has data,  
pointer to left child and a  
pointer to right child */
public class Node 
{ 
    public int data; 
    public Node left, right; 
} 
  
public class INT 
{ 
    public int v; 
    public INT(int a) 
    { 
        v = a; 
    } 
} 
  
/* utility that allocates a new  
node with the given data and  
null left and right pointers. */
public static Node newnode(int data) 
{ 
    Node node = new Node(); 
    node.data = data; 
    node.left = node.right = null; 
    return (node); 
} 
  
// function to check if there exist  
// any subtree with given sum  
// cur_sum -. sum of current subtree  
//       from ptr as root  
// sum_left -. sum of left subtree  
//           from ptr as root  
// sum_right -. sum of right subtree  
//           from ptr as root  
public static bool sumSubtreeUtil(Node ptr, INT cur_sum, int sum) 
{ 
    // base condition  
    if (ptr == null) 
    { 
        cur_sum = new INT(0); 
        return false; 
    } 
  
    // Here first we go to left  
    // sub-tree, then right subtree  
    // then first we calculate sum  
    // of all nodes of subtree having  
    // ptr as root and assign it as  
    // cur_sum. (cur_sum = sum_left +  
    // sum_right + ptr.data) after that  
    // we check if cur_sum == sum  
    INT sum_left = new INT(0), sum_right = new INT(0); 
    return (sumSubtreeUtil(ptr.left, sum_left, sum)  
            || sumSubtreeUtil(ptr.right, sum_right, sum)  
            || ((cur_sum.v = sum_left.v + sum_right.v + ptr.data) == sum)); 
} 
  
// Wrapper over sumSubtreeUtil()  
public static bool sumSubtree(Node root, int sum) 
{ 
    // Initialize sum of  
    // subtree with root  
    INT cur_sum = new INT(0); 
  
    return sumSubtreeUtil(root, cur_sum, sum); 
} 
  
// Driver Code  
public static void Main(string[] args) 
{ 
    Node root = newnode(8); 
    root.left = newnode(5); 
    root.right = newnode(4); 
    root.left.left = newnode(9); 
    root.left.right = newnode(7); 
    root.left.right.left = newnode(1); 
    root.left.right.right = newnode(12); 
    root.left.right.right.right = newnode(2); 
    root.right.right = newnode(11); 
    root.right.right.left = newnode(3); 
    int sum = 22; 
  
    if (sumSubtree(root, sum)) 
    { 
        Console.WriteLine("Yes"); 
    } 
    else
    { 
        Console.WriteLine("No"); 
    } 
} 
} 
  
// This code is contributed by Shrikant13 

Output:

Yes

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