I recently got a variation of the following Leetcode question at LinkedIn:
However, instead of multiplying by it's depth to get the sum, they wanted me to multiply by the reverseDepth which is defined as maxDepth - currentDepth - 1, i.e., if maxDepth == 3, then
In the original Leetcode problem linked above, you can typically solve it using 1 recursion.
Question: Is there a way to calculate the "reverseDepth" sum still using only 1 recursion?
I was having trouble here, so in the end, I just did 2 separate recursions
(1) recursion to calculate maxDepth
(2) recursion to calculate the reverseDepth sum (you can do this easily once you know maxDepth)