0096. 不同的二叉搜索树

题目地址

https://leetcode-cn.com/problems/unique-binary-search-trees/

题目描述

给定一个整数 n，求以 1 ... n 为节点组成的二叉搜索树有多少种？

示例:

输入: 3
输出: 5
解释:
给定 n = 3, 一共有 5 种不同结构的二叉搜索树:

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

题解

递归解法

定义递归函数 public int build(int left, int right)，用来描述通过 [left, right] 构建的所有二叉搜索树的数量。

class Solution {
    public int numTrees(int n) {
        return build(1, n);
    }

    public int build(int left, int right) {
        if (left == right) return 1;
        if (left > right) return 1; // null 子树也属于一棵树

        int numTrees = 0;
        for (int i = left; i <= right; ++ i) {
            // [left, i-1] 表示左子树
            int numLeftTrees = build(left, i-1);

            // [i+1, right] 表示右子树
            int numRightTrees = build(i+1, right);

            numTrees += numLeftTrees * numRightTrees;
        }

        return numTrees;
    }
}

耗时非常长

通过 memo 剪枝

class Solution {
    public int numTrees(int n) {
        int[][] memo = new int[n+1][n+1];
         for (int i = 0; i <= n; i++) {
            Arrays.fill(memo[i], -1);
        }

        return build(1, n, memo);
    }

    public int build(int left, int right, int[][] memo) {
        if (left == right) return 1;
        if (left > right) return 1; // null 子树也属于一棵树

        if (memo[left][right] != -1) {
            return memo[left][right];
        }

        int numTrees = 0;
        for (int i = left; i <= right; ++ i) {
            // [left, i-1] 表示左子树
            int numLeftTrees = build(left, i-1, memo);

            // [i+1, right] 表示右子树
            int numRightTrees = build(i+1, right, memo);

            numTrees += numLeftTrees * numRightTrees;
        }

        memo[left][right] = numTrees;

        return numTrees;
    }
}

执行时间瞬间下降。