95. Unique Binary Search Trees II

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95. Unique Binary Search Trees II

Description

Difficulty: Medium

Related Topics: Dynamic Programming, Backtracking, Tree, Binary Search Tree, Binary Tree

Given an integer n, return all the structurally unique **BST’**s (binary search trees), which has exactly n nodes of unique values from 1 to n. Return the answer in any order.

Example 1:

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Input: n = 3
Output: [[1,null,2,null,3],[1,null,3,2],[2,1,3],[3,1,null,null,2],[3,2,null,1]]

Example 2:

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Input: n = 1
Output: [[1]]

Constraints:

  • 1 <= n <= 8

Hints/Notes

  • Draw the tree
  • Dynamic programming

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        if (n == 0) return vector<TreeNode*>();
        return generate(1, n);
    }

    vector<TreeNode*> generate(int low, int high) {
        vector<TreeNode*> res;
        if (low > high) {
            res.push_back(nullptr);
            return res;
        } else if (low == high) {
            res.push_back(new TreeNode(low));
            return res;
        }

        for (int i = low; i <= high; i++) {
            vector<TreeNode*> leftTrees = generate(low, i - 1);
            vector<TreeNode*> rightTrees = generate(i + 1, high);
            for (TreeNode* leftTree : leftTrees) {
                for (TreeNode* rightTree : rightTrees) {
                    TreeNode* root = new TreeNode(i);
                    root->left = leftTree;
                    root->right = rightTree;
                    res.push_back(root);
                }
            }
        }
        return res;
    }
};