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39-count-no-of-subtrees-having-given-sum.cpp
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using namespace std;
// Tree Node
struct Node
{
int data;
Node* left;
Node* right;
};
// Utility function to create a new Tree Node
Node* newNode(int val)
{
Node* temp = new Node;
temp->data = val;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Function to Build Tree
Node* buildTree(string str)
{
// Corner Case
if(str.length() == 0 || str[0] == 'N')
return NULL;
// Creating vector of strings from input
// string after spliting by space
vector<string> ip;
istringstream iss(str);
for(string str; iss >> str; )
ip.push_back(str);
// Create the root of the tree
Node* root = newNode(stoi(ip[0]));
// Push the root to the queue
queue<Node*> queue;
queue.push(root);
// Starting from the second element
int i = 1;
while(!queue.empty() && i < ip.size()) {
// Get and remove the front of the queue
Node* currNode = queue.front();
queue.pop();
// Get the current node's value from the string
string currVal = ip[i];
// If the left child is not null
if(currVal != "N") {
// Create the left child for the current node
currNode->left = newNode(stoi(currVal));
// Push it to the queue
queue.push(currNode->left);
}
// For the right child
i++;
if(i >= ip.size())
break;
currVal = ip[i];
// If the right child is not null
if(currVal != "N") {
// Create the right child for the current node
currNode->right = newNode(stoi(currVal));
// Push it to the queue
queue.push(currNode->right);
}
i++;
}
return root;
}
// Your are required to complete this function
int countSubtreesWithSumX(Node* root, int x);
int main()
{
int t;
cin>>t;
getchar();
while (t--)
{
string s;
getline(cin,s);
Node* root = buildTree(s);
int x;
cin>>x;
getchar();
cout << countSubtreesWithSumX(root, x)<<endl;
}
return 0;
}
// } Driver Code Ends
//User function Template for C++
/*
Structure of the node of the binary tree is as
struct Node
{
int data;
struct Node* left;
struct Node* right;
};
*/
//Function to count number of subtrees having sum equal to given sum.
int helper(Node* curr, int& count, int& X)
{
if(curr == NULL)
{
return 0;
}
if(curr->left == NULL && curr->right == NULL)
{
if(curr->data == X)
count++;
return curr->data;
}
int ans1 = helper(curr->left, count, X);
int ans2 = helper(curr->right, count, X);
int val = ans1 + ans2 + curr->data;
if(val == X)
count++;
return val;
}
int countSubtreesWithSumX(Node* root, int X)
{
int count = 0;
helper(root, count, X);
return count;
}