Let c(x) be the number of leaves in a binary tree rooted at t. Assume that isLeaf(t) returns 1 if t is a leaf and 0 otherwise. Which of the following observations leads to a recursive implementation?

c(t)= c(t-> left)+ c(t-> right)

c(t)= c(t-> left)+ c(t-> right)+ 1

c(t)= c(t-> left)+ c(t-> right)+ isLeaf(t)

c(t)= c(t-> left)+ c(t-> right)+ isLeaf(t)+ 1

none of the above