題目地址

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

題目描述

給定一個整數(shù) n，求以 1 ... n 為節(jié)點組成的二叉搜索樹有多少種？

示例:

輸入: 3
輸出: 5
解釋:
給定 n = 3, 一共有 5 種不同結(jié)構(gòu)的二叉搜索樹:

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

題解

遞歸解法

定義遞歸函數(shù) public int build(int left, int right)，用來描述通過 [left, right] 構(gòu)建的所有二叉搜索樹的數(shù)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 子樹也屬于一棵樹

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

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

            numTrees += numLeftTrees * numRightTrees;
        }

        return numTrees;
    }
}

耗時非常長

通過 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 子樹也屬于一棵樹

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

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

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

            numTrees += numLeftTrees * numRightTrees;
        }

        memo[left][right] = numTrees;

        return numTrees;
    }
}

執(zhí)行時間瞬間下降。