[Leetcode] 95. Unique Binary Search Trees II 解题报告

题目：

Given an integer 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

思路：

1、深度优先搜索：如果要生成从1到n这n个树所构成的所有BST，按照其根节点的数值可以分为n种类型：即根节点的值为从1...n。假设我们要生成根节点为k (1 <= k <= n)的所有BST，那么首先需要生成范围为1到k-1的所有BST作为左子树，以及范围为k+1到n的所有BST作为右子树，然后两两组合就可以形成根节点为k的所有BST（注意思考这里面为什么可以保证没有重复的？）。

2、动态规划：在对深度优先搜索的思路进行分析的过程中我们发现，子问题有可能会被重复计算，从而造成效率降低。而动态规划刚好就是解决这一问题的利器！具体请见代码片段2。

3、深度优先搜索+记忆：我们可以建立一张二维表，然后存储子问题，这样在求解子问题之前，我们首先在二维表中查找，如果找到了，就可以直接返回结果，否则才开始计算。有人把这一思路叫做自顶向下的动态规划，但我觉得叫做带记忆的深度优先搜索更合适一些。

代码：

1、深度优秀搜索：

/**
* Definition for a binary tree node.
* 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<TreeNode*> tree;
if(n == 0) {
return tree;
}
dfs(1, n, tree);
return tree;
}
private:
void dfs(int start, int end, vector<TreeNode*>& tree) {
if(start > end) {
tree.push_back(NULL); // we must push a NULL value for integration
return;
}
for(int i = start; i <= end; ++i) {
// important notation: the left and right should be the copy, not the reference
vector<TreeNode*> left;
vector<TreeNode*> right;
dfs(start, i - 1, left);
dfs(i + 1, end, right);
for(int j = 0; j < left.size(); ++j) {
for(int k = 0; k < right.size(); ++k) {
TreeNode* root = new TreeNode(i);
root->left = left[j];
root->right = right[k];
tree.push_back(root);
}
}
}
}
};
2、动态规划：

/**
* Definition for a binary tree node.
* 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) {
if(n == 0) {
return vector<TreeNode*>();
}
vector<vector<vector<TreeNode*>>> dp(n + 2, vector<vector<TreeNode*>>(n + 2, vector<TreeNode*>()));
for(int i = 1; i <= n + 1; ++i) {
dp[i][i].push_back(new TreeNode(i));
dp[i][i - 1].push_back(NULL);
}
for(int l = 2; l <= n; ++l) { // length
for(int i = 1; i <= n - l + 1; ++i) { // start
for(int j = i; j <= i + l - 1; ++j) { // end
for(int k = 0; k < dp[j + 1][i + l - 1].size(); ++k) { // left subtree
for(int m = 0; m < dp[i][j - 1].size(); ++m) { // right subtree
TreeNode *T = new TreeNode(j);
T->left = dp[i][j - 1][m];
T->right = dp[j + 1][i + l - 1][k];
dp[i][i + l - 1].push_back(T);
}
}
}
}
}
return dp[1]
;
}
};3、深度优先搜索+记忆：
/**
* Definition for a binary tree node.
* 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<TreeNode*> tree;
if (n == 0) {
return tree;
}
vector<vector<vector<TreeNode*>>> dp(n,vector<vector<TreeNode*>>(n));
helper(1, n , tree, dp);
return tree;
}
private:
void helper(int start, int end, vector<TreeNode*> &tree,vector<vector<vector<TreeNode*>>> &dp) {
if (start > end) {
tree.push_back(NULL);
return;
}
if (!dp[start - 1][end - 1].empty()) {
tree = dp[start - 1][end - 1];
return;
}
for (int i = start; i <= end; ++i) {
vector<TreeNode*> left, right;
helper(start, i - 1, left, dp);
helper(i + 1, end, right, dp);
for(int j = 0; j < left.size(); ++j) {
for (int k = 0; k < right.size(); ++k) {
TreeNode* node = new TreeNode(i);
node->left = left[j];
node->right = right[k];
tree.push_back(node);
}
}
}
dp[start - 1][end - 1] = tree;
}
};