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(, 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 - );
             vector<TreeNode*> rightNode = createNode(i + , end);
             for(int j = ; j < leftNode.size(); ++j){
                 for(int k = ; 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 == )//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(, 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 - )){
                 for(TreeNode rightNode : createTree(i + , end)){
                     TreeNode node = new TreeNode(i);
                     node.left = leftNode;
                     node.right = rightNode;
                     result.add(node);
                 }
             }
         }
         return result;
     }
 }