使用 BFS 的 n-ary 树的直径

N叉树是指每个节点最多有k个子节点的有根树。 n叉树的直径是两个叶子节点之间的最长路径。

已经讨论了各种方法来计算树的直径。
1. N叉树的直径
2. O(n) 中二叉树的直径
3. 二叉树的直径
4. 使用 DFS 的树的直径

本文讨论了另一种使用 bfs 计算 n 叉树直径树的方法。

第 1 步：运行 bfs 以找到离有根树最远的节点，假设为 A
步骤 2：然后从 A 运行 bfs 以找到离 A 最远的节点让 B
第 3 步：节点 A 和 B 之间的距离是给定树的直径

C++

// C++ Program to find Diameter of n-ary tree
#include 
using namespace std;
 
// Here 10000 is maximum number of nodes in
// given tree.
int diameter[10001];
 
// The Function to do bfs traversal.
// It uses iterative approach to do bfs
// bfsUtil()
int bfs(int init, vector arr[], int n)
{
    // Initializing queue
    queue q;
    q.push(init);
 
    int visited[n + 1];
    for (int i = 0; i <= n; i++) {
        visited[i] = 0;
        diameter[i] = 0;
    }
 
    // Pushing each node in queue
    q.push(init);
 
    // Mark the traversed node visited
    visited[init] = 1;
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        for (int i = 0; i < arr[u].size(); i++) {
            if (visited[arr[u][i]] == 0) {
                visited[arr[u][i]] = 1;
 
                // Considering weight of edges equal to 1
                diameter[arr[u][i]] += diameter[u] + 1;
                q.push(arr[u][i]);
            }
        }
    }
 
    // return index of max value in diameter
    return int(max_element(diameter + 1,
                           diameter + n + 1)
               - diameter);
}
 
int findDiameter(vector arr[], int n)
{
    int init = bfs(1, arr, n);
    int val = bfs(init, arr, n);
    return diameter[val];
}
 
// Driver Code
int main()
{
    // Input number of nodes
    int n = 6;
 
    vector arr[n + 1];
 
    // Input nodes in adjacency list
    arr[1].push_back(2);
    arr[1].push_back(3);
    arr[1].push_back(6);
    arr[2].push_back(4);
    arr[2].push_back(1);
    arr[2].push_back(5);
    arr[3].push_back(1);
    arr[4].push_back(2);
    arr[5].push_back(2);
    arr[6].push_back(1);
 
    printf("Diameter of n-ary tree is %d\n",
           findDiameter(arr, n));
 
    return 0;
}

Java

// Java Program to find Diameter of n-ary tree
import java.util.*;
 
class GFG
{
     
// Here 10000 is maximum number of nodes in
// given tree.
static int diameter[] = new int[10001];
 
// The Function to do bfs traversal.
// It uses iterative approach to do bfs
// bfsUtil()
static int bfs(int init,
               Vector>arr, int n)
{
    // Initializing queue
    Queue q = new LinkedList<>();
    q.add(init);
 
    int visited[] = new int[n + 1];
    for (int i = 0; i <= n; i++)
    {
        visited[i] = 0;
        diameter[i] = 0;
    }
 
    // Pushing each node in queue
    q.add(init);
 
    // Mark the traversed node visited
    visited[init] = 1;
    while (q.size() > 0)
    {
        int u = q.peek();
        q.remove();
        for (int i = 0;
                 i < arr.get(u).size(); i++)
        {
            if (visited[arr.get(u).get(i)] == 0)
            {
                visited[arr.get(u).get(i)] = 1;
 
                // Considering weight of edges equal to 1
                diameter[arr.get(u).get(i)] += diameter[u] + 1;
                q.add(arr.get(u).get(i));
            }
        }
    }
    int in = 0;
    for(int i = 0; i <= n; i++)
    {
        if(diameter[i] > diameter[in])
        in = i;
    }
     
    // return index of max value in diameter
    return in;
}
 
static int findDiameter(Vector> arr, int n)
{
    int init = bfs(1, arr, n);
    int val = bfs(init, arr, n);
    return diameter[val];
}
 
// Driver Code
public static void main(String args[])
{
    // Input number of nodes
    int n = 6;
 
    Vector> arr = new
    Vector>();
     
    for(int i = 0; i < n + 1; i++)
    {
        arr.add(new Vector());
    }
 
    // Input nodes in adjacency list
    arr.get(1).add(2);
    arr.get(1).add(3);
    arr.get(1).add(6);
    arr.get(2).add(4);
    arr.get(2).add(1);
    arr.get(2).add(5);
    arr.get(3).add(1);
    arr.get(4).add(2);
    arr.get(5).add(2);
    arr.get(6).add(1);
 
    System.out.printf("Diameter of n-ary tree is %d\n",
                                 findDiameter(arr, n));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program to find diameter of n-ary tree
      
# Here 10000 is maximum number of nodes in
# given tree.
diameter = [0 for i in range(10001)]
  
# The Function to do bfs traversal.
# It uses iterative approach to do bfs
# bfsUtil()
def bfs(init, arr, n):
 
    # Initializing queue
    q = []
    q.append(init)
  
    visited = [0 for i in range(n + 1)]
     
    for i in range(n + 1):
        visited[i] = 0
        diameter[i] = 0
  
    # Pushing each node in queue
    q.append(init)
  
    # Mark the traversed node visited
    visited[init] = 1
     
    while (len(q) > 0):
        u = q[0]
        q.pop(0)
         
        for i in range(len(arr[u])):
            if (visited[arr[u][i]] == 0):
                visited[arr[u][i]] = 1
  
                # Considering weight of edges equal to 1
                diameter[arr[u][i]] += diameter[u] + 1
                q.append(arr[u][i])
             
    ing = 0
     
    for i in range(n + 1):
        if(diameter[i] > diameter[ing]):
            ing = i
      
    # Return index of max value in diameter
    return ing
 
def findDiameter(arr, n):
 
    init = bfs(1, arr, n)
    val = bfs(init, arr, n)
    return diameter[val]
 
# Driver Code
if __name__=='__main__':
     
    # Input number of nodes
    n = 6
  
    arr = [[] for i in range(n + 1)]
  
    # Input nodes in adjacency list
    arr[1].append(2)
    arr[1].append(3)
    arr[1].append(6)
    arr[2].append(4)
    arr[2].append(1)
    arr[2].append(5)
    arr[3].append(1)
    arr[4].append(2)
    arr[5].append(2)
    arr[6].append(1)
  
    print("Diameter of n-ary tree is " +
          str(findDiameter(arr, n)))
 
# This code is contributed by rutvik_56

C#

// C# Program to find Diameter of n-ary tree
using System;
using System.Collections.Generic;
 
class GFG
{
     
// Here 10000 is maximum number of nodes
// in given tree.
static int []diameter = new int[10001];
 
// The Function to do bfs traversal.
// It uses iterative approach to do bfs
// bfsUtil()
static int bfs(int init,
               List>arr, int n)
{
    // Initializing queue
    Queue q = new Queue();
    q.Enqueue(init);
 
    int []visited = new int[n + 1];
    for (int i = 0; i <= n; i++)
    {
        visited[i] = 0;
        diameter[i] = 0;
    }
 
    // Pushing each node in queue
    q.Enqueue(init);
 
    // Mark the traversed node visited
    visited[init] = 1;
    while (q.Count > 0)
    {
        int u = q.Peek();
        q.Dequeue();
        for (int i = 0;
                 i < arr[u].Count; i++)
        {
            if (visited[arr[u][i]] == 0)
            {
                visited[arr[u][i]] = 1;
 
                // Considering weight of edges equal to 1
                diameter[arr[u][i]] += diameter[u] + 1;
                q.Enqueue(arr[u][i]);
            }
        }
    }
    int iN = 0;
    for(int i = 0; i <= n; i++)
    {
        if(diameter[i] > diameter[iN])
        iN = i;
    }
     
    // return index of max value in diameter
    return iN;
}
 
static int findDiameter(List> arr, int n)
{
    int init = bfs(1, arr, n);
    int val = bfs(init, arr, n);
    return diameter[val];
}
 
// Driver Code
public static void Main(String []args)
{
    // Input number of nodes
    int n = 6;
 
    List> arr = new
    List>();
     
    for(int i = 0; i < n + 1; i++)
    {
        arr.Add(new List());
    }
 
    // Input nodes in adjacency list
    arr[1].Add(2);
    arr[1].Add(3);
    arr[1].Add(6);
    arr[2].Add(4);
    arr[2].Add(1);
    arr[2].Add(5);
    arr[3].Add(1);
    arr[4].Add(2);
    arr[5].Add(2);
    arr[6].Add(1);
 
    Console.Write("Diameter of n-ary tree is {0}\n",
                              findDiameter(arr, n));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
    // JavaScript Program to find Diameter of n-ary tree
     
    // Here 10000 is maximum number of nodes in
    // given tree.
    let diameter = new Array(10001);
 
    // The Function to do bfs traversal.
    // It uses iterative approach to do bfs
    // bfsUtil()
    function bfs(init, arr, n)
    {
        // Initializing queue
        let q = [];
        q.push(init);
 
        let visited = new Array(n + 1);
        for (let i = 0; i <= n; i++)
        {
            visited[i] = 0;
            diameter[i] = 0;
        }
 
        // Pushing each node in queue
        q.push(init);
 
        // Mark the traversed node visited
        visited[init] = 1;
        while (q.length > 0)
        {
            let u = q[0];
            q.shift();
            for (let i = 0;
                     i < arr[u].length; i++)
            {
                if (visited[arr[u][i]] == 0)
                {
                    visited[arr[u][i]] = 1;
 
                    // Considering weight of edges equal to 1
                    diameter[arr[u][i]] += diameter[u] + 1;
                    q.push(arr[u][i]);
                }
            }
        }
        let In = 0;
        for(let i = 0; i <= n; i++)
        {
            if(diameter[i] > diameter[In])
            In = i;
        }
 
        // return index of max value in diameter
        return In;
    }
 
    function findDiameter(arr, n)
    {
        let init = bfs(1, arr, n);
        let val = bfs(init, arr, n);
        return diameter[val];
    }
     
    // Input number of nodes
    let n = 6;
  
    let arr = [];
      
    for(let i = 0; i < n + 1; i++)
    {
        arr.push([]);
    }
  
    // Input nodes in adjacency list
    arr[1].push(2);
    arr[1].push(3);
    arr[1].push(6);
    arr[2].push(4);
    arr[2].push(1);
    arr[2].push(5);
    arr[3].push(1);
    arr[4].push(2);
    arr[5].push(2);
    arr[6].push(1);
  
    document.write("Diameter of n-ary tree is " +
    findDiameter(arr, n) + "</br>");
 

输出：

Diameter of n-ary tree is 3