Print all the paths from root, with a specified sum in Binary tree in C++

In this problem, we are given a Binary tree and a sum S. And we have to find the path starting from root to any node of the tree, which gives the sum equal to the given sum.

Input

Sum = 14
Output : path : 4 10
4 3 7

To find the solution to this problem, we need to find the preorder traversal of the binary tree. And then find the path that adds up to the given sum.

Example

#include<bits/stdc++.h>
using namespace std;
struct Node{
   int key;
   struct Node *left, *right;
};
Node* insertNode(int key){
   Node* temp = new Node;
   temp->key = key;
   temp->left = temp->right = NULL;
   return (temp);
}
void printPathsUtilSum(Node* curr_node, int sum, int
sum_so_far, vector<int> &path){
   if (curr_node == NULL)
      return;
   sum_so_far += curr_node->key;
   path.push_back(curr_node->key);
   if (sum_so_far == sum ){
      for (int i=0; i<path.size(); i++)
         cout<<path[i]<<"\t";
      cout<<endl;
   }
   if (curr_node->left != NULL)
      printPathsUtilSum(curr_node->left, sum,
   sum_so_far, path);
   if (curr_node->right != NULL)
      printPathsUtilSum(curr_node->right, sum,
   sum_so_far, path);
   path.pop_back();
}
void pathWithSum(Node *root, int sum){
   vector<int> path;
   printPathsUtilSum(root, sum, 0, path);
}
int main (){
   Node *root = insertNode(4);
   root->left = insertNode(10);
   root->right = insertNode(3);
   root->right->left = insertNode(7);
   root->right->right = insertNode(1);
   root->left->left = insertNode(8);
   root->left->right = insertNode(6);
   int sum = 14;
   cout<<"Paths with the given sum are : "<<endl;
   pathWithSum(root, sum);
   return 0;
}

Output

Paths with the given sum are −

4 10
4 3 7