Leetcode Unique Binary Search Trees

Leetcode the unique Binary Search trees original problem

Given the number of N of 1 to N, how many two-forked search trees can be formed with each of these shapes.

Note the point:

• This n number is two forks of the search tree node, can not only take part

Example:

Input: n = 3

Output: 5

   1         3     3      2      1    \       /     /      \           3     2     1      1   3      2    /     /       \                    2     1         2                 3

Thinking of solving problems

First, it is clear that the types of the two-fork search trees that have n unequal numbers are equal. and 1 to n can be used as the root node of the binary search tree, when k is the root node, it has a k-1 on the left, and its right side has n-k of unequal numbers. The type of the two-fork search tree with k as the root node is the product of the possible kinds of the left and right. A recursive representation is where h(n) = h(0)*h(n-1) + h(1)*h(n-2) + ... + h(n-1)h(0) (其中n>=2) H (0) =h (1) = 1, because 0 or 1 numbers can consist of only one shape. The value of h (x) is calculated from 1 to n in turn. In addition, this is actually a Cattleya number, can be directly calculated using mathematical formula, but the above method is more intuitive.

AC Source

 class solution(object):     def numtrees(self, n):        "" : Type N:int:rtype:int "" "DP = [1  for__inchRange (n +1)] forIinchRange2, n +1): s =0             forJinchRange (i): s + = dp[j] * dp[i-1-j] dp[i] = sreturndp[-1]if__name__ = ="__main__":assertSolution (). Numtrees (5) == the

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Leetcode Unique Binary Search Trees