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Leetcode_num6_Unique Binary Search Trees

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