LeetCode: Unique Binary Search Trees

Posted on January 20, 2018July 26, 2020 by braindenny

Unique Binary Search Trees

Similar Problems:

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: dp

## https://code.dennyzhang.com/unique-binary-search-trees
## Basic Ideas: dynamic programming
##
##    dp(i): count with i nodes
##
##     There must be one root node
##
##        dp(j)*dp(i-1-j)
##           j = 0...i-1
## Complexity: Time O(n*n), Space O(1)
class Solution:
    def numTrees(self, n: int) -> int:
        dp = [0]*(n+1)
        dp[0] = 1
        for i in range(1, n+1):
            for j in range(i):
                dp[i] += dp[j]*dp[i-1-j]
        return dp[-1]

Solution: dp

// https://code.dennyzhang.com/unique-binary-search-trees
// Basic Ideas: dynamic programming
//       Pitfalls: try to compare the values. This direction will make things very complicated
//
//       How to get f(n) from previous values?
//           1. We will have a root, so sum(left sub-tree nodes) + sum(right sub-tree nodes) = n-1
//           2. If root is i, left sub-tree will have i-1 nodes, right sub-tree will have n-k nodes.
//                  How many different types? f(i-1)*f(n-i)
//           3. Loop k from 1 to n. Then collect the total number
//  n = 2:
//
//       1         2
//        \       /
//         2     1
//
//  n = 3:
//
//       1         3     3      2      1
//        \       /     /      / \      \
//         3     2     1      1   3      2
//        /     /       \                 \
//       2     1         2                 3
//
//  n = 4:
//
//
//
// Complexity:

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

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