Path Sum

/*

Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22

*/
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void findSum(TreeNode *node, int currSum, int const sum, bool &found){
       if(node==NULL) return ;
      
       currSum = currSum+node->val;
       if(found==false&&sum==currSum&&node->left==NULL&&node->right==NULL){
          found = true;
          return;
       }
       findSum(node->left,currSum,sum,found);
       findSum(node->right,currSum,sum,found);
    }
    bool hasPathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int currSum = 0;
        bool found = false;
        findSum(root,currSum,sum,found);
        return found;
    }
};