从给定的 Prufer 序列打印每个节点的度数

给定一个 Prufer 序列，任务是找到由 prufer 序列构成的树的所有节点的度数。
例子：

Input: arr[] = {4, 1, 3, 4} 
Output: 2 1 2 3 1 1

The tree is:
2----4----3----1----5
     |
     6 

Input: arr[] = {1, 2, 2} 
Output: 2 3 1 1 1

一个简单的方法是使用 Prufer 序列创建树，然后找到所有节点的度数。
有效的方法：创建一个大小为 2 的degree[]数组，该数组的长度大于 prufer 序列的长度，因为如果N是节点数，则 prufer 序列的长度为N – 2 。最初，用1填充度数组。在 Prufer 序列中迭代并增加每个元素在度表中的频率。这种方法有效是因为 Prufer 序列中节点的频率比树中的度数小 1。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to print the degrees of every
// node in the tree made by
// the given Prufer sequence
void printDegree(int prufer[], int n)
{
    int node = n + 2;
 
    // Hash-table to mark the
    // degree of every node
    int degree[n + 2 + 1];
 
    // Initially let all the degrees be 1
    for (int i = 1; i <= node; i++)
        degree[i] = 1;
 
    // Increase the count of the degree
    for (int i = 0; i < n; i++)
        degree[prufer[i]]++;
 
    // Print the degree of every node
    for (int i = 1; i <= node; i++) {
        cout << degree[i] << " ";
    }
}
 
// Driver code
int main()
{
    int a[] = { 4, 1, 3, 4 };
    int n = sizeof(a) / sizeof(a[0]);
    printDegree(a, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    // Function to print the degrees of every
    // node in the tree made by
    // the given Prufer sequence
    static void printDegree(int prufer[], int n)
    {
        int node = n + 2;
 
        // Hash-table to mark the
        // degree of every node
        int[] degree = new int[n + 2 + 1];
 
        // Initially let all the degrees be 1
        for (int i = 1; i <= node; i++)
        {
            degree[i] = 1;
        }
 
        // Increase the count of the degree
        for (int i = 0; i < n; i++)
        {
            degree[prufer[i]]++;
        }
 
        // Print the degree of every node
        for (int i = 1; i <= node; i++)
        {
            System.out.print(degree[i] + " ");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = {4, 1, 3, 4};
        int n = a.length;
        printDegree(a, n);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
 
# Function to print the degrees of
# every node in the tree made by
# the given Prufer sequence
def printDegree(prufer, n):
  
    node = n + 2
 
    # Hash-table to mark the
    # degree of every node
    degree = [1] * (n + 2 + 1)
 
    # Increase the count of the degree
    for i in range(0, n):
        degree[prufer[i]] += 1
 
    # Print the degree of every node
    for i in range(1, node+1): 
        print(degree[i], end = " ")
      
# Driver code
if __name__ == "__main__":
  
    a = [4, 1, 3, 4]
    n = len(a)
    printDegree(a, n)
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
     
class GFG
{
 
// Function to print the degrees of every
// node in the tree made by
// the given Prufer sequence
static void printDegree(int []prufer, int n)
{
    int node = n + 2;
 
    // Hash-table to mark the
    // degree of every node
    int[] degree = new int[n + 2 + 1];
 
    // Initially let all the degrees be 1
    for (int i = 1; i <= node; i++)
    {
        degree[i] = 1;
    }
 
    // Increase the count of the degree
    for (int i = 0; i < n; i++)
    {
        degree[prufer[i]]++;
    }
 
    // Print the degree of every node
    for (int i = 1; i <= node; i++)
    {
        Console.Write(degree[i] + " ");
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = {4, 1, 3, 4};
    int n = a.Length;
    printDegree(a, n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript implementation of the approach
 
    // Function to prvar the degrees of every
    // node in the tree made by
    // the given Prufer sequence
    function printDegree(prufer , n)
    {
        var node = n + 2;
 
        // Hash-table to mark the
        // degree of every node
        var degree = Array(n + 2 + 1).fill(0);
 
        // Initially let all the degrees be 1
        for (i = 1; i <= node; i++) {
            degree[i] = 1;
        }
 
        // Increase the count of the degree
        for (i = 0; i < n; i++) {
            degree[prufer[i]]++;
        }
 
        // Print the degree of every node
        for (i = 1; i <= node; i++) {
            document.write(degree[i] + " ");
        }
    }
 
    // Driver code
     
        var a = [ 4, 1, 3, 4 ];
        var n = a.length;
        printDegree(a, n);
 
// This code contributed by Rajput-Ji
 

输出：

2 1 2 3 1 1

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