Root to leaf paths sum
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  Difficulty: Easy   Marks: 2

Given a binary tree, where every node value is a number. 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                   600+30+2=632
  6->3->5->7                6000+300+50+7=6357
  6->3->5->4                6000+300+50+4=6354
  6->5>4                    600+50+4=654   
Answer = 632 + 6357 + 6354 + 654 = 13997 

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

Your Task:
Don't take any input/output. Just complete the function treePathsSum.

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

Explanation:
Testcase2:
                
10
                /     \
             20     30
            /    \
          40    60
10*100+20*10+40*1=1240
10*100+20*10+60*1=1260
10*10+30*1=130
1240+1260+130 = 2630

** For More Input/Output Examples Use 'Expected Output' option **

Contributor: Shubham Joshi
Author: Shubham Joshi 1


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