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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
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
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