please explain clearly , what this question is asking?

For the given tree, Let c(t) 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, and that c(t) returns 0 for an empty tree. Which of the following leads to a recursive implementation of c(t)?

c(t) = isLeaf(t)

c(t) = c(t.left) + c(t.right)

c(t) = c(t.left) + c(t.right) + 1

c(t) = c(t.left) + 1 + 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

c(t) = c(t.left) + 1 + c(t.right) + 1 + isLeaf(t)

c(t) = c(t.left) + 1 + c(t.right) + 1 + isLeaf(t) + 1