Unique Binary Search Trees II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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In this post, you will find the solution for the Unique Binary Search Trees II in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either Easy, Medium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

• Mathematics/Basic Logical Based Questions
• Arrays
• Strings
• Hash Table
• Dynamic Programming
• Stack & Queue
• Trees & Graphs
• Greedy Algorithms
• Breadth-First Search
• Depth-First Search
• Sorting & Searching
• BST (Binary Search Tree)
• Database
• Linked List
• Recursion, etc.

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Link for the Problem – Unique Binary Search Trees II– LeetCode Problem

Unique Binary Search Trees II– LeetCode Problem

Problem:

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:

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:

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

Constraints:

• 1 <= n <= 8

Unique Binary Search Trees II– LeetCode Solutions

Unique Binary Search Trees II Solution in C++:

class Solution {
 public:
  vector<TreeNode*> generateTrees(int n) {
    if (n == 0)
      return {};

    return generateTrees(1, n);
  }

 private:
  vector<TreeNode*> generateTrees(int min, int max) {
    if (min > max)
      return {nullptr};

    vector<TreeNode*> ans;

    for (int i = min; i <= max; ++i)
      for (TreeNode* left : generateTrees(min, i - 1))
        for (TreeNode* right : generateTrees(i + 1, max)) {
          ans.push_back(new TreeNode(i));
          ans.back()->left = left;
          ans.back()->right = right;
        }

    return ans;
  }
};

Unique Binary Search Trees II Solution in Java:

class Solution {
  public List<TreeNode> generateTrees(int n) {
    if (n == 0)
      return new ArrayList<>();

    return generateTrees(1, n);
  }

  private List<TreeNode> generateTrees(int min, int max) {
    if (min > max)
      return Arrays.asList((TreeNode) null);

    List<TreeNode> ans = new ArrayList<>();

    for (int i = min; i <= max; ++i)
      for (TreeNode left : generateTrees(min, i - 1))
        for (TreeNode right : generateTrees(i + 1, max)) {
          ans.add(new TreeNode(i));
          ans.get(ans.size() - 1).left = left;
          ans.get(ans.size() - 1).right = right;
        }

    return ans;
  }
}

Unique Binary Search Trees II Solution in Python:

class Solution:
  def generateTrees(self, n: int) -> List[TreeNode]:
    if n == 0:
      return []

    def generateTrees(mini: int, maxi: int) -> List[Optional[int]]:
      if mini > maxi:
        return [None]

      ans = []

      for i in range(mini, maxi + 1):
        for left in generateTrees(mini, i - 1):
          for right in generateTrees(i + 1, maxi):
            ans.append(TreeNode(i))
            ans[-1].left = left
            ans[-1].right = right

      return ans

    return generateTrees(1, n)