Print All 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

 Live Demo

#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