[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][n];
    }
};

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;
    }
};