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