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

转载自http://www.tuicool.com/articles/NRjmyyI

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)

也就是说1为根时，1的左子树只能是空树，右子树是含2个节点的树；

2为根时，2的左子树是含1个节点的树，右子树也是含1个节点的树；

3为根时，左子树是含2个节点的树，右子树为空树；

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 , 150316990 @qq.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

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

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