Question

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

96	Unique Binary Search Trees

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

For example,
Given n = 3, there are a total of 5 unique BST's.

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

@tag-tree

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

Java

  • 
    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;
            }
        }
    }
    
  • // 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(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)
    
    

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