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采用Java实现两道题目；

Unique Binary Search Trees题目描述如下：

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

题目思路大致如下：对于给定的数n，自1至n分别以每个数作为节点总数计算其Unique Binary Search Trees数目result[i]，其中当节点总数为i时的result[i] 计算方法为：

左子树的Unique Binary Search Trees数目 * 右子树的Unique Binary Search Trees，所有情况相加即得result[i]，左右子树的节点个数变化范围为(0, i - 1), (1, i - 2)......(i - 1, 0), 代码如下：

public class UniqueBinarySearchTrees {      public int numTrees(int n) {          if (n == 1)              return 1;          if (n == 2)              return 2;          int[] result = new int[n + 1];          result[0] = 1;          result[1] = 1;          result[2] = 2;         for (int i = 3; i <= n; i++) {             int tmp = 0;             for (int k = 0; k < i; k++) {                 tmp += (result[k] * result[i - k - 1]);             }             result[i] = tmp;         }         return result[n];     } }

Unique Binary Search Trees II题目描述如下：

Given 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

confused what "{1,#,2,3}" means? 

题目相对上题的变化即要返回所有的Unique Binary Search Trees组成的集合，解题思路如下：以自1至n的每个数i作为根节点：比i小的所有节点组成左子树，比i大的所有节点组成右子树，迭代构造得到所有的以i为根节点的Unique Binary Search Tree；然后迭代i，即可构造出所有的Unique Binary Search Tree；代码如下：

public class Solution {    public List
   
     generateTrees(int n) {        return generateTrees(1, n);    }    public List
    
      generateTrees(int start, int end){        List
     
       result = new ArrayList
      
       ();        if(start > end){            result.add(null);            return result;        }                List
       
         leftTree = new ArrayList
        
         (); List
         
           rightTree = new ArrayList
          
           (); for(int i = start; i <= end; i++){ leftTree = generateTrees(start, i - 1); rightTree = generateTrees(i + 1, end); for(TreeNode left : leftTree) for(TreeNode right : rightTree){ TreeNode root = new TreeNode(i); root.right = right; root.left = left; result.add(root); } } return result; }}
          
         
        
       
      
     
    
   

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