LEETCODE —— Unique Binary Search Trees [动态规划]

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

 Tree Dynamic Programming

 

 

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