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 + , vector<TreeNode *>());
if(n == )
{
ans[].push_back(NULL);
return ans[];
} TreeNode * root = NULL;
ans[].push_back(root);
root = new TreeNode();
ans[].push_back(root);
for(int i = ; i <= n; i++) //总数字
{
for(int j = ; j < i; j++) //小于根节点的数字个数
{
for(int l = ; l < ans[j].size(); l++) //小于根节点的组成方法数
{
for(int r = ; r < ans[i - j - ].size(); r++) //大于根节点的组成方法数
{
TreeNode * root = new TreeNode(j + );
root->left = ans[j][l];
root->right = add(ans[i - j - ][r], j + ); //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(); return ;
}
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 - );
vector<TreeNode*> right = generateTreesRec(i + , end);
TreeNode *node;
for(int j = ; j < left.size(); ++j){
for(int k = ; 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(, n);
}
};