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

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++ program to find if there is a subtree with
// given sum
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";
        cout << "No";
    return 0;



This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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