Question

Formatted question description: https://leetcode.ca/all/96.html

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

Algorithm

Let’s first look at the situation when n = 1, only a single binary search tree can be formed. The situations where n is 1, 2, and 3 are as follows:

			1                        n = 1



                2        1                   n = 2
               /          \
              1            2



   1         3     3      2      1           n = 3
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

We assign 1 when n = 0, because the empty tree is also a binary search tree, then the situation when n = 1 can be regarded as the number of left subtrees multiplied by the number of right subtrees, and the left and right subtrees Trees are all empty trees, so 1 by 1 is still 1. Then when n = 2, since both 1 and 2 can be roots, you can calculate them separately and add them together. The case of n = 2 can be calculated by the following formula (where dp[i] represents the number of BST that can be composed of i numbers):

dp[2] = 

	dp[0] * dp[1]　　　 
	(If 1 is the root, the left subtree must not exist, and the right subtree can have a number)

	+ 

	dp[1] * dp[0]　　 
	(If 2 is the root, the left subtree can have a number, and the right subtree must not exist)

In the same way, the calculation method for n = 3 can be written:

dp[3] = 
	dp[0] * dp[2]　　　 
	(If 1 is the root, the left subtree must not exist, and the right subtree can have two numbers)

	+ 
	
	dp[1] * dp[1]　　 
	(If 2 is the root, the left and right subtrees can each have a number)

 　  + 
 
 	dp[2] * dp[0]　　 
	(If 3 is the root, the left subtree can have two numbers, and the right subtree must not exist)

Code

public class Unique_Binary_Search_Trees {

    public static void main(String[] args) {
        Unique_Binary_Search_Trees out = new Unique_Binary_Search_Trees();
        Solution s = out.new Solution();

        System.out.println(s.numTrees(3));
    }

    public class Solution {
        public int numTrees(int n) {
            if (n <= 0) {
                return 0;
            }

            // dp[i] represents the number of BST that can be composed of i numbers
            int[] dp = new int[n + 1];
            dp[0] = 1; // null is counted as one unique tree

            for (int i = 1; i <= n; i++) { // 1...n

                for (int j = 1; j <= i; j++) {
                    // for each fixed n, calculate its sum
                    dp[i] += dp[j - 1] * dp[i - j];
                }
            }

            return dp[n];
        }
    }

    public class Solution_recursion {
        public int numTrees(int n) {
            if (n < 0) {
                return 0;
            }

            if (n == 0 || n == 1) {
                return 1;
            }

            int sum = 0;

            for (int i = 1; i <= n; i++) {

                // 2 conditions: unique && BST. => inorder-visit will generate ordered sequence
                // so, if decide root, then left and right can be calculated
                // @note: root is "i", left has "i-1" nodes, right has "n - (i - 1) - 1"=="n - i" nodes
                sum += numTrees(i - 1) * numTrees(n - i);
            }

            return sum;
        }
    }
}

############

class Solution {
    public int numTrees(int n) {
        int[] dp = new int[n + 1];
        dp[0] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < i; ++j) {
                dp[i] += dp[j] * dp[i - j - 1];
            }
        }
        return dp[n];
    }
}

// OJ: https://leetcode.com/problems/unique-binary-search-trees
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int numTrees(int n) {
        int dp[20] = {[0 ... 1] = 1};
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j <= i; ++j) {
                dp[i] += dp[j - 1] * dp[i - j];
            }
        }
        return dp[n];
    }
};

class Solution:
    def numTrees(self, n: int) -> int:
        dp = [0] * (n + 1)
        dp[0] = 1 # e.g, null is counted as one unique tree
        for i in range(1, n + 1):
            # up to i-1, when j=i-1 meaning left child having i-1 node then i is the root
            for j in range(i):
                dp[i] += dp[j] * dp[i - j - 1] # -1, not counting root
        return dp[-1]

############

class Solution(object):
  def _numTrees(self, n):
    """
    :type n: int
    :rtype: int
    """
    dp = [0] * (n + 1)
    dp[0] = dp[1] = 1
    for i in range(2, n + 1):
      for j in range(1, i + 1):
        dp[i] += dp[j - 1] * dp[i - j]
    return dp[-1]

  def numTrees(self, n):
    ans = 1
    for i in range(1, n + 1):
      ans = ans * (n + i) / i
    return ans / (n + 1)

func numTrees(n int) int {
	dp := make([]int, n+1)
	dp[0] = 1
	for i := 1; i <= n; i++ {
		for j := 0; j < i; j++ {
			dp[i] += dp[j] * dp[i-j-1]
		}
	}
	return dp[n]
}

96 - Unique Binary Search Trees

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