# Tree Recursion

In the previous examples we saw recursive functions that call themselves one time within the code. This type of recursion is called linear recursion, where head and tail recursion are two specific types of linear recursion.

In this section we will investigate another type of recursion called tree recursion, which occurs when a function calls itself two or more times to solve a single problem. To illustrate tree recursion, we will use a simple recursive function MAX, which finds the maximum of $N$ elements in an array. To calculate the maximum of $N$ elements we will use the following recursive algorithm.

1. Compute the maximum of the first $N/2$ elements and store in MAX1.
2. Compute the maximum of the last $N/2$ elements and store in MAX2.
3. Compare MAX1 and MAX2 to find the maximum of all elements.

Our process recursively decomposes the problem by searching for the maximum in the first $N/2$ elements and the second $N/2$ elements until we reach the base case. In this problem, the base case is when we either have 1 or 2 elements in the array. If we just have 1, we return that value. If we have 2, we return the larger of those two values. An overview of the process is shown below.

The pseudocode for the algorithm is shown below.

function MAX(VALUES, START, END)
print "Called MAX with start = " + START + ", end = " + END
if END – START = 0
return VALUES[START]
else if END – START = 1
if VALUES(START) > VALUES(END)
return VALUES[START]
else
return VALUES[END]
end if
else
MIDDLE = ROUND((END – START) / 2)
MAX1 = MAX(VALUES, START, START + MIDDLE – 1)
MAX2 = MAX(VALUES, START + MIDDLE, END)
if MAX1 > MAX2
return MAX1
else
return MAX2
end if
end if
end function

The following block shows the output from the print line in the MAX function above. The initial call to the function is MAX(VALUES, 0, 15).

Called MAX with start = 0, end = 7
Called MAX with start = 0, end = 3
Called MAX with start = 0, end = 1
Called MAX with start = 2, end = 3
Called MAX with start = 4, end = 7
Called MAX with start = 4, end = 5
Called MAX with start = 6, end = 7
Called MAX with start = 8, end = 15
Called MAX with start = 8, end = 11
Called MAX with start = 8, end = 9
Called MAX with start = 10, end = 11
Called MAX with start = 12, end = 15
Called MAX with start = 12, end = 13
Called MAX with start = 14, end = 15

As you can see, MAX decomposes the array each time it is called, resulting in 14 instances of the MAX function being called. If we had performed head or tail recursion to compare each value in the array, we would have to have called MAX 16 times. While this may not seem like a huge savings, as the value of $N$ grows, so do the savings.