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LeetCode "Unique Binary Search Trees"
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; }};
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