LeetCode OJ ：Unique Binary Search Trees II（唯一二叉搜索树）

题目如下所示：返回的结果是一个Node的Vector：

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
　　
树节点的定义是下面这样的 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        return createNode(1, n);
    }
    
    vector<TreeNode*> createNode(int start, int end)
    {
        vector<TreeNode*> result;
        if(start > end){
            result.push_back(NULL);
            return result;
        }
        for(int i = start; i <= end; ++i){
            vector<TreeNode*> leftNode = createNode(start, i - 1);
            vector<TreeNode*> rightNode = createNode(i + 1, end);
            for(int j = 0; j < leftNode.size(); ++j){
                for(int k = 0; k < rightNode.size(); ++k){
                    TreeNode * tmpNode = new TreeNode(i);
                    tmpNode->left = leftNode[j];　　
                    tmpNode->right = rightNode[k];
                    result.push_back(tmpNode);
                }
            }
        }
        return result;
    }
};

这一题实际上更另外一个叫做different ways to add parentheses的题目比较相似，这个详见上一篇博文。

java版本的代码如下：

/**
 * 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> generateTrees(int n) {
        if(n == 0)//如果不加上这一条，当为0的时候会返回[[]]，不知道为什么，很奇怪
            return new ArrayList<TreeNode>();
        return createTree(1, n);
    }

    public List<TreeNode> createTree(int start, int end){
        ArrayList<TreeNode> result = new ArrayList<TreeNode>();
        if(start > end){
            result.add(null);
            return result;
        }
        for(int i = start; i <= end; ++i){
            for(TreeNode leftNode : createTree(start, i - 1)){
                for(TreeNode rightNode : createTree(i + 1, end)){
                    TreeNode node = new TreeNode(i);
                    node.left = leftNode;
                    node.right = rightNode;
                    result.add(node);
                }
            }
        }
        return result;
    }
}