95. Unique Binary Search Trees II

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

Topic: Given the value n, return all possible BSTs consisting of 1. n. Note that a BST only needs to return a root pointer.

Ideas: See also https://leetcode.com/problems/unique-binary-search-trees-ii/discuss/31494/A-simple-recursive-solution . For BST with root node i, the left subtree left contains elements of start..i-1 and right subtree right contains elements of I + 1.. End. Assuming that lefts is a set of all possible left subtrees (each subtree has only a head node), right is a set of all right subtrees, then for BST with root i, only lefts and HTS need to be combined. The root nodes at different locations in the range of [start,end] are recursively iterated.

Same thinking LeetCode241. Different Ways to Add Parentheses.

Engineering Code Download

/**
 * 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 < 1) return vector<TreeNode*>{};
        return generateTrees(1, n);
    }
private:
    vector<TreeNode*> generateTrees(int start, int end){
        vector<TreeNode*> res;

        if(start > end){
            res.push_back(nullptr);
            return res;
        }

        if(start == end){
            TreeNode* root = new TreeNode(start);
            res.push_back(root);
            return res;
        }

        //In the range of `[start, end]', for the case where root node is i
        for(int i = start; i <= end; ++i){
            //Notice the left boundary of left and the right boundary of right.
            vector<TreeNode*> left = generateTrees(start, i-1);
            vector<TreeNode*> right = generateTrees(i+1, end);

            for(auto l : left){
                for(auto r : right){
                    TreeNode* root = new TreeNode(i);
                    root->left = l;
                    root->right = r;
                    res.push_back(root);
                }
            }
        }

        return res;
    }
};