[LeetCode] Unique Binary Search Trees II dfs 深度搜索

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.

 

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  Tree Dynamic Programming

 

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　　这个嘛，对于1 to n ，如果要用某个值做节点，那么这个值左部分的全部可能的树，递归调用获得，右部分同理，这样便可以获取结果。

 

#include <iostream>

#include <vector>

using namespace std;



/**

 * 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 help_f(1,n);

    }

    vector<TreeNode *> help_f(int l,int r)

    {

        vector<TreeNode *> ret;

        if(l>r){

            ret.push_back(NULL);

            return ret;

        }

        for(int i=l;i<=r;i++){

            vector<TreeNode *> lPart = help_f(l,i-1);

            vector<TreeNode *> rPart = help_f(i+1,r);

            for(int lidx=0;lidx<lPart.size();lidx++){

                for(int ridx=0;ridx<rPart.size();ridx++){

                    TreeNode * pNode = new TreeNode(i);

                    pNode->left = lPart[lidx];

                    pNode->right = rPart[ridx];

                    ret.push_back(pNode);

                }

            }

        }

        return ret;

    }

};



int main()

{

    return 0;

}