【Leetcode】Unique Binary Search Trees

假設給定n個節點，節點值爲1,2,3...,n，求由這些結點可以構成多少棵不同的二叉查找樹。

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

思路：遞歸，由於是二叉查找樹，先選擇任一結點根結點，假設爲結點i，則[1，i-1]範圍的結點爲結點i的左子樹結點，[i+1，n]範圍的結點爲結點i的右子樹結點，則以結點i爲根結點的BST個數爲左，右子樹可構成BST個數的乘積，基於這個思路，可以寫出以下遞歸程序。

class Solution {
public:
	int numTrees(int n) 
	{
		return numTrees(1,n);
	}

	int numTrees(int start, int end)
	{
		if (start >= end)
			return 1;

		int totalNum = 0;
		for (int i=start; i<=end; ++i)
			totalNum += numTrees(start,i-1)*numTrees(i+1,end);
		return totalNum;
	}
};