leetcode95 Unique Binary Search Trees II

题目：

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.

思路：

本题采取递归的思路。

传递的参数是开始数值(begin)和结束数值(end)。

当begin > end 时，返回空(注意不是null)；

当begin == end 时， 返回含有 new TreeNode(begin)结点的ArrayList；

当begin < end时，建立两个ArrayList来分别接收左右子树。

代码：

     public  List<TreeNode> generateTrees(int n){
         return generateTrees(1, n);
     }

     public List<TreeNode> generateTrees(int begin, int end){
         List<TreeNode> arr = new ArrayList<TreeNode>();
         if(begin > end){
             return arr;
         }
         if(begin == end){
             TreeNode ptr = new TreeNode(begin);
             arr.add(ptr);
             return arr;
         }
         for(int i = begin; i <= end; i++){
             List<TreeNode> left = new ArrayList<TreeNode>();
             List<TreeNode> right = new ArrayList<TreeNode>();
             left = generateTrees(begin, i-1);
             right = generateTrees(i+1, end);
             //注意判断left和right是否为空
             //还有，要注意应该在最内层循环每次都新建根结点
             if(left.size() == 0){
                 if(right.size() == 0){
                     TreeNode root = new TreeNode(i);
                     root.left = null;
                     root.right = null;
                     arr.add(root);
                 }else{
                     for(TreeNode r: right){
                         TreeNode ptr = new TreeNode(i);
                         ptr.left = null;
                         ptr.right = r;
                         arr.add(ptr);
                     }
                 }
             }else{
                 if(right.size() == 0){
                     for(TreeNode l: left){
                         TreeNode ptr = new TreeNode(i);
                         ptr.left = l;
                         ptr.right = null;
                         arr.add(ptr);
                     }
                 }else{
                     for(TreeNode l: left){
                         for(TreeNode r: right){
                             TreeNode ptr = new TreeNode(i);
                             ptr.left = l;
                             ptr.right = r;
                             arr.add(ptr);
                         }
                     }
                 }
             }
         }
         return arr;
     }