题目来源:https://leetcode.com/problems/unique-binary-search-trees/
问题描述
96. Unique Binary Search Trees
Medium
Given n, how many structurally unique BST's (binary search trees) that store values 1 ... n?
Example:
Input: 3
Output: 5
Explanation:
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
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题意
给定整数n,求所有n个节点的二叉搜索树的数量(给定二叉搜索树的顺序为“左子树<根节点<右子树”)
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思路
动态规划。动态规划数组的dp[i]表示i个节点的二叉搜索树的个数。考虑n个节点的情形,其中必有一个为根节点,剩下n-1个节点可以有0/1/2/…/n-1位于左子树,n-1/n-2/n-3/…/0个位于右子树,因此状态转移方程为:
dp[n] = dp[0]*dp[n-1] + dp[1]*dp[n-2] + dp[2]*dp[n-3] + … + dp[n-1]*dp[0]
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代码
class Solution {
public int numTrees(int n) {
int[] dp = new int[n+1];
dp[0] = 1;
dp[1] = 1;
for (int i=2; i<=n; i++)
{
for (int j=0; j<i; j++)
{
dp[i] += dp[j] * dp[i-1-j];
}
}
return dp[n];
}
}