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;
         }
     }
 }