[96] leetcode different binary search tree (dynamic programming, binary search tree)

Topic links: https://leetcode-cn.com/problems/unique-binary-search-trees/

Title Description

Given an integer n, 1 ... n is seeking to nodes of a binary search tree, how many?

Example:

输入: 3
输出: 5
解释:
给定 n = 3, 一共有 5 种不同结构的二叉搜索树:

   1         3     3      2      1
    \       /     /      / \      \
     3     2     1      1   3      2
    /     /       \                 \
   2     1         2                 3

Thinking

First, because the current integer iwhen the inevitable situation and <=icircumstances related to, so we consider the idea of dynamic programming, first establish a one-dimensional dparray.
dp[i]He expressed as an integer inumber of binary search tree species composition of the time.
We consider i as the root node, set up sub-tree nodes is left j, 0<=j<=n-1then the right child tree nodes i-1-j. The total number of binary search tree is composed of the combined total of the left subtree and right subtree . That dynamic programming equation.

dp[i] = sum( dp[j] ,dp[i-1-j)) for j in [0,n-1]

The resulting formula is the number of Cattleya

G(n) = G(0) * G(n-1) + G(1) * G(n-2) + … + G(n-1) * G(0) 

Complexity Analysis
time complexity: O (n ^ 2)
Complexity Space: O (1)

Code

class Solution {
public:
    int numTrees(int n) {
        vector<int> dp(n+1,0);
        dp[0] = 1, dp[1] = 1;
        for(int i = 2; i<= n; ++i)
            for (int j = 0; j <= i-1; ++j)
                dp[i] += (dp[j] * dp[i-1-j]);
        return dp[n];
    }
};