Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
For example,
Given n = 3, there are a total of 5 unique BST's.1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3
如果集合为空，只有一种BST，即空树， UniqueTrees[0] =1 如果集合仅有一个元素，只有一种BST，即为单个节点 UniqueTrees[1] = 1
UniqueTrees[2] = UniqueTrees[0] * UniqueTrees[1] (1为根的情况)
+ UniqueTrees[1] * UniqueTrees[0] (2为根的情况。再看一遍三个元素的数组，可以发现BST的取值方式如下：UniqueTrees[3] = UniqueTrees[0]*UniqueTrees[2] (1为根的情况) + UniqueTrees[1]*UniqueTrees[1] (2为根的情况) + UniqueTrees[2]*UniqueTrees[0] (3为根的情况)所以，由此观察，可以得出UniqueTrees的递推公式为UniqueTrees[i] = ∑ UniqueTrees[0...k] * [i-1-k] k取值范围 0<= k <=(i-1) '''
Created on Nov 13, 2014
@author: ScottGu<gu.kai.66@gmail.com, kai.gu@live.com>
'''
class Solution :
# @return an integer
def numTrees( self , n):
uniqueTrees={}
uniqueTrees[ 0 ]= 1
uniqueTrees[ 1 ]= 1
for cnt in range( 2 , n+ 1 ):
uniqueTrees[cnt]= 0
for k in range( 0 , cnt):
uniqueTrees[cnt]+=uniqueTrees[k]*uniqueTrees[cnt- 1 -k]
return uniqueTrees[n]
if __name__ == '__main__' :
sl=Solution()
print sl.numTrees( 0 ), 0
print sl.numTrees( 1 ), 1
print sl.numTrees( 2 ), 2
print sl.numTrees( 3 ), 3
print sl.numTrees( 4 ), 4
print sl.numTrees( 5 ), 5