Unique Binary Search Trees LeetCode Solution | Easy Approach

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Given an integer n, return the number of structurally unique BST’s (binary search trees) which has exactly n nodes of unique values from 1 to n.

Example 1:

Input: n = 3
Output: 5

Example 2:

Input: n = 1
Output: 1

Constraints:

  • 1 <= n <= 19

 Unique Binary Search Trees Solutions

Time: O(n)
Space: O(n)

C++

class Solution {
 public:
  int numTrees(int n) {
    // G[i] := # of unique BST's that store values 1..i
    vector<int> G(n + 1);
    G[0] = 1;
    G[1] = 1;

    for (int i = 2; i <= n; ++i)
      for (int j = 0; j < i; ++j)
        G[i] += G[j] * G[i - j - 1];

    return G[n];
  }
};

Java

 class Solution {
  public int numTrees(int n) {
    // G[i] := # of unique BST's that store values 1..i
    int[] G = new int[n + 1];
    G[0] = 1;
    G[1] = 1;

    for (int i = 2; i <= n; ++i)
      for (int j = 0; j < i; ++j)
        G[i] += G[j] * G[i - j - 1];

    return G[n];
  }
}

Python

class Solution:
  def numTrees(self, n: int) -> int:
    # G[i] := # of unique BST's that store values 1..i
    G = [1, 1] + [0] * (n - 1)

    for i in range(2, n + 1):
      for j in range(i):
        G[i] += G[j] * G[i - j - 1]

    return G[n]

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