Root to leaf paths sum

PROBLEM :
Given a binary tree, where every node value is a Digit from 1-9 . Find the sum of all the numbers which are formed from root to leaf paths.

For example consider the following Binary Tree.


           6                              
         /   \                        
        3     5                    
      /   \     \
     2    5      4            
        /  \                      
       7    4                
           
  There are 4 leaves, hence 4 root to leaf paths:
  Path                      Number
  6->3->2                   632
  6->3->5->7                6357
  6->3->5->4                6354
  6->5>4                    654  
Answer = 632 + 6357 + 6354 + 654 = 13997
Input:
The task is to complete the method which takes one argument, root of Binary Tree. The Node has a data part which stores the data, pointer to left child and pointer to right child. There are multiple test cases. For each test case, this method will be called individually.

Output:
The function should return sum of all the numbers which are formed from root to leaf paths.

Constraints:
1 <=T<= 30
1 <=Number of nodes<= 100
1 <=Data of a node<= 1000

Example:
Input:
2
2
1 2 L 1 3 R
4
10 20 L 10 30 R 20 40 L 20 60 R

Output:
25
2630

There are two test cases.  First case represents a tree with 3 nodes and 2 edges where root is 1, left child of 1 is 2 and right child of 1 is 3. Second test case represents a tree with 4 edges and 5 nodes.

--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------

/* Tree node structure  used in the program
 struct Node
 {
     int data;
     Node* left, *right;
}; */

/*You are required to complete below method */

void inorderTraversal(Node *,long long [],int *,long long ) ;
long long treePathsSum(Node *root)
{
    int i ;
    long long arr[100] ;
    i=0 ;
    long long sum,ele=0 ;
   
    inorderTraversal(root,arr,&i,ele) ;
   
    sum=0 ;
    for(int j=0;j<i;j++)
        sum+=arr[j] ;
   
    return sum ;
}

void inorderTraversal(Node *root,long long arr[],int *i,long long ele)
{
    if(root==NULL)
        return ;
       
    ele=ele*10+root->data ;  
    if(root->left==NULL&&root->right==NULL)
    {
        arr[*i]=ele ;
        (*i)++ ;
        return ;
    }
   
    inorderTraversal(root->left,arr,&(*i),ele) ;
    inorderTraversal(root->right,arr,&(*i),ele) ;
}

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