96. Unique Binary Search Trees
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题目:
Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
For example,
Given n = 3, there are a total of 5 unique BST's.
1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3题解:
一组单调递增的值,能构成多少棵结构上独一无二的二叉搜索树,只和这组值的个数有关,和值无关。
根据原题得
状态转移方程:
g(n) = g(0)*g(n-1) + g(1) * g(n-2) + g(2) * g(n-3) + ......+g(n-1) * g(0).
具体代码如下:
class Solution {public: int numTrees(int n) { int * g = new int[n+1]; g[0] = g[1] = 1; for(int i = 2; i <= n; i++) { g[i] = 0; for(int j = 1; j <= i; j++) { g[i] += (g[j-1] * g[i-j]); } } return g[n]; }};end!
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- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
- 96. Unique Binary Search Trees
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