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LeetCode "Unique Binary Search Trees"

时间:2014-07-18 18:18:01      阅读:187      评论:0      收藏:0      [点我收藏+]

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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;
    }
};

LeetCode "Unique Binary Search Trees",布布扣,bubuko.com

LeetCode "Unique Binary Search Trees"

标签:style   blog   color   io   for   re   

原文地址:http://www.cnblogs.com/tonix/p/3853414.html

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