96. Unique Binary Search Trees

Problem statement:

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

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

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

Solution:

The problem wants the number of unique BST tree. DP will solve it.

dp[i] means how many bst trees it can build if n == i. The size of dp array is n + 1.

The initial status is dp[0] = 1. There is only one solution when n = 0.

If there are i tree nodes, i - 1 nodes in total for left and right tree. We enumerate all possible solutions from 0 to i - 1, and return dp[n]

The value of dp[n] = dp[j] * dp[n - 1 -j] (0 <= j < n),

j: means how many nodes in left, n - 1 - j: how many nodes in right tree

class Solution {
public:
    int numTrees(int n) {
        vector<int> dp(n + 1, 0);
        dp[0] = 1;
        for(int i = 1; i <= n; i++){
            for(int j = 0; j < i; j++){
                // if there are j nodes in left tree.
                // the number of nodes in right tree is i - 1 - j
                dp[i] += dp[i - 1 - j] * dp[j];                    
            }
        }
        return dp[n];
    }
};