先简单介绍下两个函数的区别，treeAggregate在aggregate的基础上会多一个depth参数，默认为2，另外在执行comb函数时，aggregate会多执行一次，目前实验没看到其他区别。
aggregate有三个参数，长的很奇怪，是吧？第一个参数是zeroValue，可以理解为初始值，第二个参数是seq函数，第三个参数是comb函数。treeAggregate类似，只是多了一个depth参数。
seq函数作用于每个元素，comb函数作用于每个seq函数输出的结果。
测试代码如下：
val d = sc.parallelize(List(1, 2),3)
def seq(a:Int, b:Int):Int={
val result = a * b * b
println("seq, " + a + ", " + b + ", " + result)
result
}
def comb(a:Int, b:Int):Int={
val result = a + b
println("comb, " + a + ", " + b + ", " + result)
result
}
d.aggregate(100)(seq, comb)
d.treeAggregate(100)(seq, comb, 2)
d.treeAggregate(100)(seq, comb, 3)输出如下：
# d.aggregate(100)(seq, comb) seq, 100, 2, 400 seq, 100, 1, 100 comb, 100, 400, 500 comb, 500, 100, 600 comb, 600, 100, 700 Int = 700
# d.treeAggregate(100)(seq, comb, 2) seq, 100, 2, 400 seq, 100, 1, 100 comb, 100, 100, 200 comb, 200, 400, 600 Int = 600
# d.treeAggregate(100)(seq, comb, 3) seq, 100, 1, 100 seq, 100, 2, 400 comb, 100, 400, 500 comb, 500, 100, 600 Int = 600
看点更复杂的例子，求1到20的平方和：
val d = sc.parallelize(1 to 20, 5)
def seq(a:(Int, Int), b:Int):(Int, Int)={
val sum = a._1 + b * b
val count = a._2 + 1
println("seq:\t" + a + "\t" + b + "\t" + sum + "\t" + count)
(sum, count)
}
def comb(a:(Int, Int), b:(Int, Int)):(Int, Int)={
val sum = a._1 + b._1
val count = a._2 + b._2
println("comb:\t" + a + "\t" + b + "\t" + sum + "\t" + count)
(sum, count)
}
d.aggregate((0,0))(seq, comb)
d.treeAggregate((0,0))(seq, comb, 2)
d.treeAggregate((0,0))(seq, comb, 3)