Unique Binary Search Trees,Unique Binary Search Trees2 生成二叉排序树

Unique Binary Search Trees：求生成二叉排序树的个数。

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

 算法分析：类似上阶梯，简单的动态规划问题。当根节点为i时，比i小的节点有i-1个，比i大的节点有n-i个，所以，i为根节点能够生成二叉排序树的个数是

nums[n] += nums[i-1]*nums[n-i]，i从1到n。

public class UniqueBinarySearchTrees
{
	public int numTrees(int n)
	{
        if(n <= 0)
        {
        	return 0;
        }
        int[] res = new int[n+1];
        res[0] = 1;
        res[1] = 1;
        for(int i = 2; i <= n; i ++)
        {
        	for(int j = 1; j <= i; j ++)//j为根节点
        	{
        		res[i] += res[j-1]*res[i-j];
        	}
        }
        return res[n];
    }
}

 Unique Binary Search Trees2：求生成二叉排序树的根节点的集合

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.

For example,
Given n = 3, your program should return all 5 unique BST's shown below.

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

算法分析：这个不是求个数，而是求生成树根节点。使用递归。

public class UniqueBinarySearchTreesII 
{
	public List<TreeNode> generateTrees(int n) 
	{
		if(n <= 0)
		{
			return new ArrayList<TreeNode>();
		}
		
		return helper(1, n);
    }
	
	public List<TreeNode> helper(int m, int n)
	{
		List<TreeNode> res = new ArrayList<>();
		if(m > n)
		{
			res.add(null);
			return res;
		}
		
		for(int i = m; i <= n; i ++)
		{
			//i为根节点
			List<TreeNode> ls = helper(m, i-1);//i节点的左子树
			List<TreeNode> rs = helper(i+1, n);//i节点的右子树
			for(TreeNode l : ls)
			{
				for(TreeNode r : rs)
				{
					TreeNode curr = new TreeNode(i);
					curr.left = l;
					curr.right = r;
					res.add(curr);
				}
			}
		}
		return res;
	}
}