bst 刪除節點

Problem statement: C++ program to find number of binary search trees with n nodes.

問題陳述： C ++程序查找具有n個節點的二進制搜索樹的數量。

Input format: single integer n

輸入格式：單整數n

Constraints: 0=<n<=15

約束： 0 = <n <= 15

Sample input: 4

樣本輸入： 4

Sample output: 14 binary search tree/trees are there for 4 nodes

輸出示例： 14個二叉搜索樹/四個樹的樹

Problem explanation:

問題說明：

The number of BSTs with n vertices is given by Catalan numbers. For n=0,1,2,3,4... Catalan numbers are 1,1,2,5,14... and so on.

具有n個頂點的BST的數量由加泰羅尼亞語數字給出。 對于n = 0,1,2,3,4 ...加泰羅尼亞語數字是1,1,2,5,14 ...依此類推。

Catalan numbers are given by Cn = (2n)!/(n+1)!*n! = count of BSTs with nodes n.

加泰羅尼亞數字由Cn =(2n)！/(n + 1)！* n！ =節點為n的BST的計數 。

Catalan numbers are used here to find the count of BSTs because both satisfy same recurrence relation that is:

由于兩者都滿足相同的遞歸關系，因此此處使用加泰羅尼亞語數字查找BST的計數：

For n=0 number of trees is 1 i.e. empty tree. For subsequent values:

對于n = 0 ，樹的數量為1，即空樹。 對于后續值：

And, so on...

等等...

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Solution:

解：

If we consider root as the i^th node then:

如果我們將root視為第i ^個節點，則：

i-1 nodes are there in left subtree.

i-1節點在左子樹中。

n-i nodes are there in right subtree.

ni個節點在右子樹中。

Let’s denote count of BST by Bn for n elements

我們用n表示Bn的BST計數

The 2 subtrees here will be independent of each other. Therefore it will be ( B i-1 * B n-i ) for Bi . For n nodes (as i has n choices) it will be :

這里的2個子樹將彼此獨立。 因此，Bi將為(B i-1 * B ni)。 對于n個節點(因為我有n個選擇)，它將為：

Since the recurrence relation is same as of catalan numbers , so count of BST is given by Cn.

由于遞歸關系與加泰羅尼亞數相同，因此BST的計數由Cn給出。

Recurrence relation:

遞歸關系：

This gives complexity O(4^n). Complexity can be reduced to O(n^2) by using DP.

這給出了復雜度O(4 ^ n)。 使用DP可以將復雜度降低到O(n ^ 2)。

C++ implementation:

C ++實現：

#include <iostream> using namespace std;int CN(int n){int Cn =0;// base caseif(n==0) // empty tree{return 1;}for(int i=1;i<n+1;i++){Cn+= CN(i-1)*CN(n-i);}return Cn; }int main(){int n;cout<<"Enter number of nodes: ";cin>>n;cout<<n<<endl;int trees=CN(n);cout<<trees<<" binary search trees are there for "<<n<<" nodes"<<endl;return 0; }

Output

輸出量

Enter number of nodes: 4 14 binary search trees are there for 4 nodes

翻譯自: https://www.includehelp.com/cpp-programs/find-number-of-bsts-with-n-nodes-catalan-numbers.aspx

bst 刪除節點

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