[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;
}