[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(1,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-1);
            vector<TreeNode *> rTree=creatTre(k+1,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=0;i<lTree.size();++i)
                for(int j=0;j<rTree.size();++j)
                {
                    TreeNode *root=new TreeNode(k);
                    root->left=lTree[i];
                    root->right=rTree[j];
                    res.push_back(root);
                }
        }
        return res;
    }
};