Given a binary tree and an integer S, check whether there is root to leaf path with its sum as S.

Example 1:

Input:
Tree = 
            1
          /   \
        2      3
S = 2

Output: 0

Explanation:
There is no root to leaf path with sum 2.

Example 2:

Input:
Tree = 
            1
          /   \
        2      3
S = 4

Output: 1

Explanation:
The sum of path from leaf node 3 to root 1 is 4.

Your Task:
You dont need to read input or print anything. Complete the function hasPathSum() which takes root node and target sum S as input parameter and returns true if path exists otherwise it returns false.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(height of tree)

Constraints:
1 ≤ N ≤ 10^4
1 ≤ S ≤ 10^6