[Leetcode] Unique binary search trees ii 唯一二叉搜索树

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

题中很重要的一点是，各个结点的值是递增的，即非降型。所以，对任一点，其前的点构成二叉树的左子树，其后点构成二叉树右子树。可以从这两部分中随意的选取一部分组成子二叉树的左右子树。递归方法的思路是，遍历这n个数，然后从当前数的前后部分选取、组合成新的二叉树。终止条件是beg>end。感觉很哄哄的思想-水中的鱼的博客，用指针及堆来存储变量。

 /**
  * Definition for binary tree
  * 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 creatTre(,n);
     }

     vector<TreeNode *> creatTre(int beg,int end)
     {
         vector<TreeNode *> res;
         if(beg>end)
         {
             res.push_back(NULL);
             return res;
         }
         for(int k=beg;k<=end;++k)
         {
             //求根节点i的左右子树集合
             vector<TreeNode *> lTree=creatTre(beg,k-);
             vector<TreeNode *> rTree=creatTre(k+,end);

             /*将左右子树相互匹配，每一个左子树都与所有右子树匹配,
             *每一个右子树都与所有的左子树匹配  */
             for(int i=;i<lTree.size();++i)
                 for(int j=;j<rTree.size();++j)
                 {
                     TreeNode *root=new TreeNode(k);
                     root->left=lTree[i];
                     root->right=rTree[j];
                     res.push_back(root);
                 }
         }
         return res;
     }
 };