LeetCode "Unique Binary Search Trees"

最新推荐文章于 2020-02-29 22:29:24 发布

转载最新推荐文章于 2020-02-29 22:29:24 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/tonix/p/3853414.html

本文讨论了一个使用递归方法解决的问题，即计算具有特定根节点的子树数量。通过实例展示如何将问题分解为更小的子问题，并通过循环计算不同子树的数量来得出最终结果。

Another recursion-style problem.

1. Each count of sub-solution with a certain root should contribute to final count

2. With a certain root, the num of left child sub-tree multiply the right one, is the current count

1A, YEAH!

class Solution {
public:

    int numTrees(int n) {
        if (n <= 1) return 1;
        if (n == 2) return 2;
        int cnt = 0;
        for (int i = 1; i <= n; i++)
        {
            int r1 = numTrees(i - 1);
            int r2 = numTrees(n - i);
            cnt += r1 * r2;
        }
        return cnt;
    }
};