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;
    }