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

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