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 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          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
          
            generateTrees(int n) { vector
           
            
             > ans(n + 1, vector
             
              ()); 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
              
                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
     
       generateTreesRec(int start, int end){        vector
      
        v;        if(start > end){            v.push_back(NULL);            return v;        }        for(int i = start; i <= end; ++i){            vector
       
         left = generateTreesRec(start, i - 1);            vector
        
          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
         
           generateTrees(int n) {        return generateTreesRec(1, n);    }};