LeetCode – Unique Binary Search Trees (Java)

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

Analysis

Let count[i] be the number of unique binary search trees for i. The number of trees are determined by the number of subtrees which have different root node. For example,

i=0, count[0]=1 //empty tree

i=1, count[1]=1 //one tree

i=2, count[2]=count[0]*count[1] // 0 is root
            + count[1]*count[0] // 1 is root

i=3, count[3]=count[0]*count[2] // 1 is root
            + count[1]*count[1] // 2 is root
            + count[2]*count[0] // 3 is root

i=4, count[4]=count[0]*count[3] // 1 is root
            + count[1]*count[2] // 2 is root
            + count[2]*count[1] // 3 is root
            + count[3]*count[0] // 4 is root
..
..
..

i=n, count[n] = sum(count[0..k]*count[k+1...n]) 0 <= k < n-1

Use dynamic programming to solve the problem.

Java Solution

public int numTrees(int n) {
	int[] count = new int[n + 1];
 
	count[0] = 1;
	count[1] = 1;
 
	for (int i = 2; i <= n; i++) {
		for (int j = 0; j <= i - 1; j++) {
			count[i] = count[i] + count[j] * count[i - j - 1];
		}
	}
 
	return count[n];
}

Check out how to get all unique binary search trees.

Category >> Algorithms >> Interview  
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
<pre><code> 
String foo = "bar";
</code></pre>
  • Amay Yadav

    (2n)!/(n+1)!n!

  • Aditya Vutukuri

    Catalan number of ways

  • Zhiyong Tan

    Is there a typo here?

    i=2, count[2]=count[0]*count[1] // 0 is root

    + count[1]*count[0] // 1 is root

    There is no node with val 0, so 0 cannot be root. Comments should be // 1 is root
    // 2 is root.