【leetcode】 Unique Binary Search Trees II (middle)☆

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

这次的题目要求是得到所有的树。

 

我的思路：

用f[n]存储1-n的所有方法的根节点

则 f[n+1] = 1作为根，f[0]做左子树，f[n]所有节点都加1做右子树  +  2作为根，f[1]做左子树，f[n - 1]所有节点都加2做右子树 +...

代码内存都没有释放，不过AC了。

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
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) {
        vector<vector<TreeNode *>> ans(n + 1, vector<TreeNode *>());
        if(n == 0)
        {
            ans[0].push_back(NULL);
            return ans[0];
        }
        
        TreeNode * root = NULL;
        ans[0].push_back(root);
        root = new TreeNode(1);
        ans[1].push_back(root);
        for(int i = 2; i <= n; i++) //总数字
        {
            for(int j = 0; j < i; j++) //小于根节点的数字个数
            {
                for(int l = 0; l < ans[j].size(); l++) //小于根节点的组成方法数
                {
                    for(int r = 0; r < ans[i - j - 1].size(); r++) //大于根节点的组成方法数
                    {
                        TreeNode * root = new TreeNode(j + 1);
                        root->left = ans[j][l]; 
                        root->right = add(ans[i - j - 1][r], j + 1); //大于根节点的需要加上差值
                        ans[i].push_back(root);
                    }
                }
            }
        }

        return ans[n];
    }

    TreeNode * add(TreeNode * root, int Num)
    {
        if(root == NULL)
        {
            return root;
        }
        TreeNode * T = new TreeNode(root->val + Num);
        T->left = add(root->left, Num);
        T->right = add(root->right, Num);

        return T;
    }
};

int main()
{
    Solution s;
    vector<TreeNode *> ans = s.generateTrees(3);

    return 0;
}

 

看别人的思路，把建立子树分为从数字start-end，直接递归求解，不需要像我的那样每次还要把右子树遍历增加值

/**
 * 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*> generateTreesRec(int start, int end){
        vector<TreeNode*> v;
        if(start > end){
            v.push_back(NULL);
            return v;
        }
        for(int i = start; i <= end; ++i){
            vector<TreeNode*> left = generateTreesRec(start, i - 1);
            vector<TreeNode*> right = generateTreesRec(i + 1, end);
            TreeNode *node;
            for(int j = 0; j < left.size(); ++j){
                for(int k = 0; k < right.size(); ++k){
                    node = new TreeNode(i);
                    node->left = left[j];
                    node->right = right[k];
                    v.push_back(node);
                }
            }
        }
        return v;
    }
    vector<TreeNode *> generateTrees(int n) {
        return generateTreesRec(1, n);
    }
};