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 = 22 Output: "Yes" // sum of all nodes of subtree {1, 7, 12, 2} = 22 Input : sum = 15 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 #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; }

Output:

Yes

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