LeetCode OJ 95. Unique Binary Search Trees II

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? > read more on how binary tree is serialized on OJ.

这道题目要求构建出存储1...n的所有平衡二叉树，并且这些树是相互独立的，如果有M棵树，那么每个节点都要被new M次。我的想法就是遍历1~n，构建出每一个数字作为根节点时它的左子树和右子树，然后根节点加所有左右子树的组合就是我们要构建的BST。很明显这是递归的思想，代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<TreeNode> list = new ArrayList();
    public List<TreeNode> generateTrees(int n) {
        if(n<=0) return list;
        list = trees(1, n);
        return list;
    }
    
    public List<TreeNode> trees(int n, int m){  //构建BST
        List<TreeNode> clist = new ArrayList();
        if(m<n){
            clist.add(null);
            return clist;
        }
        if(m==n){
            TreeNode node = new TreeNode(n);
            clist.add(node);
            return clist;
        }
        for(int i = n; i<=m; i++){
            List<TreeNode> llist = trees(n, i-1);//构建所有的左子树
            List<TreeNode> rlist = trees(i+1, m);//构建所有的右子树
            for(TreeNode lnode:llist){           //生成以i为根节点的所有BST
                for(TreeNode rnode:rlist){
                    TreeNode root = new TreeNode(i);
                    root.left = copyTree(lnode);
                    root.right = copyTree(rnode);
                    clist.add(root);
                }
            }
        }
        return clist;
    }
    
    public TreeNode copyTree(TreeNode root){    //复制二叉树
        if(root==null) return null;
        else{
            TreeNode croot = new TreeNode(root.val);
            croot.left = copyTree(root.left);
            croot.right = copyTree(root.right);
            return croot;
        }
    }
}