[Leetcode]Unique binary search tree
2015-01-21 14:17
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1. 没有元素，为1 count[0] = 1
2. 1个元素，为1 count[1] = 1
3. 2个元素，1为根加上2为根
count[2] = count[0] * count[1] + count[1] * count[0]
推理得到
count[3] = count[0] * count[2] + count[1] * count[1] + count[2] * count[0]
写的时候注意循环的条件。
2. 1个元素，为1 count[1] = 1
3. 2个元素，1为根加上2为根
count[2] = count[0] * count[1] + count[1] * count[0]
推理得到
count[3] = count[0] * count[2] + count[1] * count[1] + count[2] * count[0]
写的时候注意循环的条件。
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