Update 437_path_sum_III.md

Author: Joseph Luce

leetcode/easy/437_path_sum_III.md

@@ -16,7 +16,7 @@ For each recursive call, we want to traverse the left and right children down to
 Since we would have to do this again but for the left and right children, it about O(log(N)) for each.
 So we basically have O(N), O(log(N)), O(log(N)).
 The two O(log(N)) can be simplified to O(N-1).
-So O(N) and O(N-1) per call = O(2^N)
+So O(N) and O(N-1) per call = O(2^N).
 
 Remember the formula for recursion, (Number of calls ^ (Big O per call))