Total number of possible Binary Search Trees using Catalan Number
Given an integer N, the task is to count the number of possible Binary Search Trees with N keys.
Input: N = 2 Output: 2 For N = 2, there are 2 unique BSTs 1 2 \ / 2 1 Input: N = 9 Output: 4862
Approach: The number of binary search trees that will be formed with N keys can be calculated by simply evaluating the corresponding number in Catalan Number series.
First few Catalan numbers for n = 0, 1, 2, 3, … are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …
Catalan numbers satisfy the following recursive formula:
Below is the implementation of the above approach:
The problem can be solved in a dynamic programming way.
Here is a snippet of how the recurrence tree will proceed:
G(4) / | | \ G(0)G(3) G(1)G(2) G(2)G(1) G(3)G(0) / | \ G(0)G(2) G(1)G(1) G(2)G(0) / \ G(0)G(1) G(1)G(0) // base case
Note: Without memoization, the time complexity is upper bounded by O(N x N!).
Given a sequence 1…n, to construct a Binary Search Tree (BST) out of the sequence, we could enumerate each number i in the sequence, and use the number as the root, naturally, the subsequence 1…(i-1) on its left side would lay on the left branch of the root, and similarly the right subsequence (i+1)…n lay on the right branch of the root. We then can construct the subtree from the subsequence recursively. Through the above approach, we could ensure that the BST that we construct is all unique since they have unique roots.
The problem is to calculate the number of unique BST. To do so, we need to define two functions:
1.G(n): the number of unique BST for a sequence of length n. 2.F(i, n), 1 <= i <= n: The number of unique BST, where the number i is the root of BST, and the sequence ranges from 1 to n. As one can see, G(n) is the actual function we need to calculate in order to solve the problem. And G(n) can be derived from F(i, n), which at the end, would recursively refer to G(n). First of all, given the above definitions, we can see that the total number of unique BST G(n), is the sum of BST F(i) using each number i as a root. i.e., G(n) = F(1, n) + F(2, n) + ... + F(n, n). Given a sequence 1…n, we pick a number i out of the sequence as the root, then the number of unique BST with the specified root F(i), is the cartesian product of the number of BST for its left and right subtrees.For example, F(2, 4): the number of unique BST tree with number 2 as its root. To construct an unique BST out of the entire sequence [1, 2, 3, 4] with 2 as the root, which is to say, we need to construct an unique BST out of its left subsequence  and another BST out of the right subsequence [3,4], and then combine them together (i.e. cartesian product). F(i, n) = G(i-1) * G(n-i) 1 <= i <= n Combining the above two formulas, we obtain the recursive formula for G(n). i.e. G(n) = G(0) * G(n-1) + G(1) * G(n-2) + … + G(n-1) * G(0)
In terms of calculation, we need to start with the lower number, since the value of G(n)
depends on the values of G(0) … G(n-1).
Below is the above implementation of the above algorithm:
Time Complexity: O(N2)
Space Complexity: O(N)
In this post, we will discuss an O(n) and an O(1) space solution based on Dynamic Programming.
We know that the formula for Catalan number for a variable n is which simplifies to
Similarily Catalan number for (n-1) nodes =
The formula for n nodes can be rewritten as
= Catalan number for (n-1) nodes*
So for every iteration for ‘i’ going from 1 to n we will store catalan number for ‘i-1’ nodes and compute for ith node.
Below is the implementation for the above approach:
Number of Unique BST for 4 nodes is 14
Time Complexity: O(n)
Auxiliary Space: O(1).