题目连接:Leetcode 095 Unique Binary Search Trees II
解题思路:dfs,参数包括数值范围start,end,枚举当前结点的值i,然后递归计算左子树start~i-1,右子树i+1~end。然后将左右子树组合到一起。
/**
* 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*> generateTree(int start, int end) {
vector<TreeNode*> ret;
if (start > end) {
ret.push_back(0);
return ret;
}
for (int i = start; i <= end; i++) {
vector<TreeNode*> l = generateTree(start, i-1);
vector<TreeNode*> r = generateTree(i+1, end);
for (int x = 0; x < l.size(); x++) {
for (int y = 0; y < r.size(); y++) {
TreeNode* c = new TreeNode(i);
c->left = l[x];
c->right = r[y];
ret.push_back(c);
}
}
}
return ret;
}
vector<TreeNode*> generateTrees(int n) {
if (n == 0) return {};
return generateTree(1, n);
}
};