[leetcode] 95. Unique Binary Search Trees II 解题报告
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题目链接:https://leetcode.com/problems/unique-binary-search-trees-ii/
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
思路: 递归构造左右子树。比较困难的是如何存储每一棵树。递归的产生每一个子树的左右节点,构造当前节点,最后可以的得到完整的一棵树。当前i节点的左节点可以是(1,i-1)的所有节点,同样右节点可以是(i+1, n)的所有节点,因此枚举所有的左右节点的可能值,最终得到所有的树。
具体代码如下:
/** * 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*> DFS(int n, int k1, int k2) { if(k1 > k2) return vector<TreeNode*>{NULL}; vector<TreeNode*> result; for(int i = k1; i <= k2; i++) { auto vec1 = DFS(n, k1, i-1), vec2 = DFS(n, i+1, k2); for(auto val1: vec1) { for(auto val2: vec2) { TreeNode *root = new TreeNode(i); root->left = val1, root->right = val2; result.push_back(root); } } } return result; } vector<TreeNode*> generateTrees(int n) { if(n <=0) return {}; return DFS(n, 1, n); }};
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