96. Unique Binary Search Trees

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class Solution {
    public int numTrees(int n) {
        // 二叉搜索树，关键在于枚举 root 节点，确定 root 节点后，小于 root 节点值的在左边，大于 root 节点值的在右边

        int[] dp = new int[n + 1];
        dp[0] = dp[1] = 1;

        for (int i = 2; i < dp.length; i++) {
            for (int root = 1; root <= i; root++) {
                dp[i] += dp[root - 1] * dp[i - root]; // cartesian product
            }
        }

        return dp[n];
    }
}

References

96. Unique Binary Search Trees