LeetCode-Unique Binary Search Trees（JAVA）

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

Example:

Input: 3 Output: 5 Explanation: 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

思路：在n的二叉搜索樹中中，假設i爲定點，則0~ i-1在i的左側，i+1~n在i的右側，以i爲頂點的二叉搜索樹的個數爲 左側i個數的個數*右側（n-i-1）的個數

class Solution {
    Map<String, Integer> maps = new HashMap<>();
    public int numTrees(int n) {
        //return tree(1, n);
        return tree2(n);
    }
    public int tree2(int n) {
        if(n < 2) {
            return 1;
        }
        int[] t = new int[n+1];
        t[0] = 1;
        t[1] = 1;
        t[2] = 2;
        
        for(int i = 3; i < n + 1 ; i++) {
            for(int j = 1; j < i+1; j++) {
                t[i] += t[j-1] * t[i-j];
            }
            
        }
        return t[n];
        
    }
    public int tree(int l, int r) {
        String temp = String.valueOf(l) + "_" + String.valueOf(r);
        if(maps.containsKey(temp)) {
            return maps.get(temp);
        }
        if(l == r) {
            return 1;
        }
        if(l > r) {
            return 0;
        }
        int num = 0;
        //num += tree(l+1,r);
        for(int i = l; i < r + 1; i++) {
            int l1 = tree(l,i - 1);
            int r1 = tree(i+1,r);
            if(l1 == 0) {
                num += r1;
            } else if(r1 == 0) {
                num += l1;
            } else {
                num = num + r1*l1;
            }
        }
        maps.put(temp, num);
        return num;
    }
    

}