Maximum value at each level in an N-ary Tree

Given a N-ary Tree consisting of nodes valued in the range [0, N – 1] and an array arr[] where each node i is associated to value arr[i], the task is to print the maximum value associated with any node at each level of the given N-ary Tree.

Examples:

Input: N = 8, Edges[][] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}}, 
arr[] = {4, 2, 3, -5, -1, 3, -2, 6} 
Output: 4 3 6 
Explanation:
Below is the given N-ary Tree: 

The Max of all nodes of the 0^th level is 4. 
The Max of all nodes of the 1^st level is 3. 
The Max of all nodes of the 2^nd level is 6.

Input: N = 10, Edges[][] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}, {6, 8}, {6, 9}}, 
arr[] = {1, 2, -1, 3, 4, 5, 8, 6, 12, 7} 
Output: 1 3 8 12 
Explanation:
Below is the given N-ary Tree: 

The Max of all nodes of the 0^th level is 1. 
The Max of all nodes of the 1^st level is 3. 
The Max of all nodes of the 2^nd level is 8. 
The Max of all nodes of the 3^rd level is 12. 

Approach: This problem can be solved by performing the Level Order Traversal of the given tree. While traversing the tree, process nodes of each level separately. For every level being processed, compute the maximum value of all nodes in the level. Follow the steps below: 

1. Store all the child nodes of the current level in a Queue and pop the nodes of the current level one by one.
2. Find the maximum value of all the popped nodes of the current level.
3. Print the maximum value obtained in the above step.
4. Follow the above steps for each level of the given Tree and print the maximum value for each level respectively.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum value
// at each level of N-ary tree
int maxAtLevel(int N, int M,
               vector<int> Value,
               int Edges[][2])
{
    // Stores the adjacency list
    vector<int> adj[N];

    // Create the adjacency list
    for (int i = 0; i < M; i++) {
        int u = Edges[i][0];
        int v = Edges[i][1];
        adj[u].push_back(v);
    }

    // Perform level order traversal
    // of nodes at each level
    queue<int> q;

    // Push the root node
    q.push(0);

    // Iterate until queue is empty
    while (!q.empty()) {

        // Get the size of queue
        int count = q.size();

        int maxVal = 0;

        // Iterate for all the nodes
        // in the queue currently
        while (count--) {

            // Dequeue an node from queue
            int temp = q.front();
            q.pop();

            maxVal = max(maxVal,
                         Value[temp]);

            // Enqueue the children of
            // dequeued node
            for (int i = 0;
                 i < adj[temp].size();
                 i++) {
                q.push(adj[temp][i]);
            }
        }

        // Print the result
        cout << maxVal << " ";
    }
}

// Driver Code
int main()
{
    // Number of nodes
    int N = 10;

    // Edges of the N-ary tree
    int Edges[][2] = { { 0, 1 }, { 0, 2 },
                       { 0, 3 }, { 1, 4 },
                       { 1, 5 }, { 3, 6 },
                       { 6, 7 }, { 6, 8 },
                       { 6, 9 } };

    // Given cost
    vector<int> Value = { 1, 2, -1, 3, 4,
                          5, 8, 6, 12, 7 };

    // Function Call
    maxAtLevel(N, N - 1, Value, Edges);

    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{

// Function to find the maximum value
// at each level of N-ary tree
static void maxAtLevel(int N, int M,
                       int []Value,
                       int Edges[][])
{
  // Stores the adjacency list
  Vector<Integer> []adj = new Vector[N];

  for (int i = 0; i < adj.length; i++)
    adj[i] = new Vector<Integer>();

  // Create the adjacency list
  for (int i = 0; i < M; i++)
  {
    int u = Edges[i][0];
    int v = Edges[i][1];
    adj[u].add(v);
  }

  // Perform level order traversal
  // of nodes at each level
  Queue<Integer> q = new LinkedList<>();

  // Push the root node
  q.add(0);

  // Iterate until queue is empty
  while (!q.isEmpty())
  {
    // Get the size of queue
    int count = q.size();

    int maxVal = 0;

    // Iterate for all the nodes
    // in the queue currently
    while (count-- > 0)
    {
      // Dequeue an node from queue
      int temp = q.peek();
      q.remove();

      maxVal = Math.max(maxVal, Value[temp]);

      // Enqueue the children of
      // dequeued node
      for (int i = 0;
               i < adj[temp].size(); i++)
      {
        q.add(adj[temp].get(i));
      }
    }

    // Print the result
    System.out.print(maxVal + " ");
  }
}

// Driver Code
public static void main(String[] args)
{
  // Number of nodes
  int N = 10;

  // Edges of the N-ary tree
  int Edges[][] = {{0, 1}, {0, 2},
                   {0, 3}, {1, 4},
                   {1, 5}, {3, 6},
                   {6, 7}, {6, 8},
                   {6, 9}};

  // Given cost
  int []Value = {1, 2, -1, 3, 4,
                 5, 8, 6, 12, 7};

  // Function Call
  maxAtLevel(N, N - 1, Value, Edges);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach

# Function to find the maximum value
# at each level of N-ary tree
def maxAtLevel(N, M, Value, Edges):

    # Stores the adjacency list
    adj = [[] for i in range(N)]

    # Create the adjacency list
    for i in range(M):
        u = Edges[i][0]
        v = Edges[i][1]
        adj[u].append(v)

    # Perform level order traversal
    # of nodes at each level
    q = []

    # Push the root node
    q.append(0)

    # Iterate until queue is empty
    while (len(q)):

        # Get the size of queue
        count = len(q)

        maxVal = 0

        # Iterate for: all the nodes
        # in the queue currently
        while (count):

            # Dequeue an node from queue
            temp = q[0]
            q.remove(q[0])

            maxVal = max(maxVal, Value[temp])

            # Enqueue the children of
            # dequeued node
            for i in range(len(adj[temp])):
                q.append(adj[temp][i])

            count -= 1

        # Print the result
        print(maxVal, end = " ")

# Driver Code
if __name__ == '__main__':

    # Number of nodes
    N = 10

    # Edges of the N-ary tree
    Edges = [ [ 0, 1 ], [ 0, 2 ],
              [ 0, 3 ], [ 1, 4 ],
              [ 1, 5 ], [ 3, 6 ],
              [ 6, 7 ], [ 6, 8 ],
              [ 6, 9 ] ]

    # Given cost
    Value = [ 1, 2, -1, 3, 4,
              5, 8, 6, 12, 7 ]

    # Function Call
    maxAtLevel(N, N - 1, Value, Edges)

# This code is contributed by ipg2016107

C#

// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{

// Function to find the
// maximum value at each
// level of N-ary tree
static void maxAtLevel(int N, int M,
                       int []Value,
                       int [,]Edges)
{
  // Stores the adjacency list
  List<int> []adj = new List<int>[N];

  for (int i = 0; i < adj.Length; i++)
    adj[i] = new List<int>();

  // Create the adjacency list
  for (int i = 0; i < M; i++)
  {
    int u = Edges[i, 0];
    int v = Edges[i, 1];
    adj[u].Add(v);
  }

  // Perform level order traversal
  // of nodes at each level
  Queue<int> q = new Queue<int>();

  // Push the root node
  q.Enqueue(0);

  // Iterate until queue is empty
  while (q.Count != 0)
  {
    // Get the size of queue
    int count = q.Count;

    int maxVal = 0;

    // Iterate for all the nodes
    // in the queue currently
    while (count-- > 0)
    {
      // Dequeue an node from queue
      int temp = q.Peek();
      q.Dequeue();

      maxVal = Math.Max(maxVal,
                        Value[temp]);

      // Enqueue the children of
      // dequeued node
      for (int i = 0;
               i < adj[temp].Count; i++)
      {
        q.Enqueue(adj[temp][i]);
      }
    }

    // Print the result
    Console.Write(maxVal + " ");
  }
}

// Driver Code
public static void Main(String[] args)
{
  // Number of nodes
  int N = 10;

  // Edges of the N-ary tree
  int [,]Edges = {{0, 1}, {0, 2},
                  {0, 3}, {1, 4},
                  {1, 5}, {3, 6},
                  {6, 7}, {6, 8},
                  {6, 9}};

  // Given cost
  int []Value = {1, 2, -1, 3, 4,
                 5, 8, 6, 12, 7};

  // Function Call
  maxAtLevel(N, N - 1, Value, Edges);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program for the above approach

// Function to find the maximum value
// at each level of N-ary tree
function maxAtLevel(N, M, Value, Edges)
{

    // Stores the adjacency list
    let adj = new Array(N);

    for(let i = 0; i < adj.length; i++)
        adj[i] = [];

    // Create the adjacency list
    for(let i = 0; i < M; i++)
    {
        let u = Edges[i][0];
        let v = Edges[i][1];
        adj[u].push(v);
    }

    // Perform level order traversal
    // of nodes at each level
    let q = [];

    // Push the root node
    q.push(0);

    // Iterate until queue is empty
    while (q.length > 0)
    {

        // Get the size of queue
        let count = q.length;

        let maxVal = 0;

        // Iterate for all the nodes
        // in the queue currently
        while (count-- > 0)
        {

            // Dequeue an node from queue
            let temp = q[0];
            q.shift();

            maxVal = Math.max(maxVal, Value[temp]);

            // Enqueue the children of
            // dequeued node
            for(let i = 0; i < adj[temp].length; i++)
            {
                q.push(adj[temp][i]);
            }
        }

        // Print the result
        document.write(maxVal + " ");
    }
}

// Driver code

// Number of nodes
let N = 10;

// Edges of the N-ary tree
let Edges = [ [ 0, 1 ], [ 0, 2 ],
              [ 0, 3 ], [ 1, 4 ],
              [ 1, 5 ], [ 3, 6 ],
              [ 6, 7 ], [ 6, 8 ],
              [ 6, 9 ] ];

// Given cost
let Value = [ 1, 2, -1, 3, 4,
              5, 8, 6, 12, 7 ];

// Function Call
maxAtLevel(N, N - 1, Value, Edges);

// This code is contributed by suresh07

</script>

Output:

1 3 8 12

Time Complexity: O(N)
Auxiliary Space: O(N)