【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
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
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
;
}

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);
}
};