regular leetcode

Author: KrisYu

096._unique_binary_search_trees.md

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+###96. Unique Binary Search Trees
+
+
+
+题目:
+<https://leetcode.com/problems/unique-binary-search-trees/>
+
+
+难度:
+Medium
+
+思路：
+
+
+参照此处hint:
+
+<https://shenjie1993.gitbooks.io/leetcode-python/content/096%20Unique%20Binary%20Search%20Trees.html>
+
+
+首先明确n个不等的数它们能构成的二叉搜索树的种类都是相等的.
+
+毫无头绪，对于1...n的bst，可以这样看，k可以作为root，那么1..k-1必定在左边，k+1...n必定在右边，而1...k-1课产生的bst树是dp[k-1],右边产生的数是dp[n-k],所以能生成的树的数量是 dp[k-1]* dp[n-k]
+
+dp[n] = sum(dp[k-1]*dp[n-k])，从0到k
+
+
+```
+class Solution(object):
+    def numTrees(self, n):
+        """
+        :type n: int
+        :rtype: int
+        """
+        dp = [ 1 for i in range(n+1)]
+
+        for i in range(2,n+1):
+        	s = 0
+        	for k in range(i):
+        		s += dp[k]*dp[i-k-1]
+        	dp[i] = s 
+
+        return dp[-1]
+```