[leetcode]95. Unique Binary Search Trees II
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题目链接:https://leetcode.com/problems/unique-binary-search-trees-ii/#/description
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:由于1~n是升序列,因此建起来的树天然就是BST。可以看出这也是一个可以划分成子问题求解的题目,所以考点是动态规划。
但具体对于本题来说,采取的是自底向上的求解过程。
1. 选出根结点后应该先分别求解该根的左右子树集合,也就是根的左子树有若干种,它们组成左子树集合,根的右子树有若干种,它们组成右子树集合。
2. 然后将左右子树相互配对,每一个左子树都与所有右子树匹配,每一个右子树都与所有的左子树匹配。然后将两个子树插在根结点上。
3. 最后,把根结点放入链表中。
class Solution {public: vector<TreeNode *> generateTrees(int n) { if(n<1) return vector<TreeNode*>(); return Helper(1, n); } vector<TreeNode *> Helper(int begin, int end) { vector<TreeNode *> ret; if(begin > end) ret.push_back(NULL); else if(begin == end) { TreeNode* node = new TreeNode(begin); ret.push_back(node); } else { for(int i = begin; i <= end; i ++) {//root vector<TreeNode *> left = Helper(begin, i-1); vector<TreeNode *> right = Helper(i+1, end); for(int l = 0; l < left.size(); l ++) { for(int r = 0; r < right.size(); r ++) { //new tree TreeNode* root = new TreeNode(i); root->left = left[l]; root->right = right[r]; ret.push_back(root); } } } } return ret; }};
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