95. Unique Binary Search Trees II

Problem:

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1 ... n.

Example:

Input: 3
Output:
[
  [1,null,3,2],
  [3,2,null,1],
  [3,1,null,null,2],
  [2,1,3],
  [1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST's shown below:

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

思路：

同样采用递归的思想。若节点\(i\)作为根节点，则只需分别构造\(1, \cdots, i-1\)的左子树和\(i+1, \cdots, n\)的右子树。同样的，左右子树的构造也是一个递归的过程。

注意：vector<TreeNode*>数组中储存的是根节点的值，所以要将根和左子树及右子树连起来时，只需分别将存储左右子树的vector<TreeNode*>left, right中的元素遍历一遍，作为根的左右子节点即构造完毕。

Solution (C++):

public:
    vector<TreeNode*> generateTrees(int n) {
        if (n == 0)  return vector<TreeNode*> {};
        return genTrees(1, n);
    }
private:
    vector<TreeNode*> genTrees(int start, int end) {
        vector<TreeNode*> bst;
        vector<TreeNode*> left, right;
        
        if (start > end)  {
            bst.push_back(NULL);
            return bst;
        }
        if (start == end) {
            TreeNode* s = new TreeNode(start);
            bst.push_back(s);
            return bst;
        }
        
        for (int i = start; i <= end; ++i) {    //以i为根
            left = genTrees(start, i-1);
            right = genTrees(i+1, end);
            for (auto lnode: left) {
                for (auto rnode: right) {
                    TreeNode* root = new TreeNode(i);
                    root->left = lnode;
                    root->right = rnode;
                    bst.push_back(root);
                }
            }
        }
        return bst;
    }

性能：

Runtime: 20 ms  Memory Usage: 19.1 MB