题目:
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
做此题还是挺顺利哒,两遍就AC啦,给自己点个赞!刷题打卡中~~~~
初看此题的思路是要用到递归,所以关键点在如何由numTrees(n-1)转化为numTrees(n),即找递推公式即可
首先将BST(n)拆成BST(n-1)和节点n,通过观察unique BSTs可知 左子树值<节点值<右子树值
I 节点n作为根节点,BST(n-1)作为左子树
II BST(n-1)作为根节点,节点n作为右子树
III 节点n插入BST(n-1),即BST(i)为根节点,n为BST(i)右子树,BST(n-1-i)为n的左子树
综上所述,可得递推公式
最后上代码啦
class Solution: # @return an integer def numTrees(self, n): if n<=0: return 0 elif n==1: return 1 elif n==2: return 2 else: rs=0 for i in range(1,n-1): rs+=self.numTrees(i)*self.numTrees(n-1-i) rs+=2*self.numTrees(n-1) return rs
Leetcode_num6_Unique Binary Search Trees
原文地址:http://blog.csdn.net/eliza1130/article/details/39478475