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Print List of nodes of given n-ary Tree with number of children in range [0, n]

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  • Difficulty Level : Basic
  • Last Updated : 31 Jan, 2023
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Given an n-ary Tree having N nodes numbered from 1 to N, the task is to print a list of nodes containing 0, 1, 2, 3, . . ., n children.

Note: An n-ary Tree is a tree in which nodes can have at most n children.

Examples:

Input:

Output
0 child: 3, 5, 6, 7
1 child: 4
2 child: 2
3 child: 1

Input:

Output
0 child: 15, 30, 40, 99, 22, 46, 56, 25, 90
1 child: 34, 60, 70
2 child: 12, 21
3 child: 50
5 child: 20

 

Approach: The idea to solve the problem by iterating the tree using level order traversal.

Follow the steps mentioned below to solve the problem:

  • Traverse from i = 1 to N:
    • For each node use level order traversal to find the location of that value in the tree.
    • Upon finding that node, use the list of children and calculate the number of children it has (say X).
    • Use a map to add ith node in the list of nodes that have X children. 
  • Iterate the map and print the nodes having the number of children in the range [0 to n].

Below is the implementation of the above approach : 

C++




// C++ code for the above approach:
 
#include <bits/stdc++.h>
using namespace std;
 
// Node Class
class Node {
public:
    int key_ele;
    vector<Node*> child;
 
    Node(int data_value)
    {
        key_ele = data_value;
    }
};
 
// Function to find number of child
int numberOfChild(Node* root, int ele)
{
    int num = 0;
 
    if (root == NULL) {
        return 0;
    }
 
    // Initializing queue data structure
    queue<Node*> q;
    q.push(root);
 
    while (!q.empty()) {
        int n = q.size();
        while (n > 0) {
            Node* nn = q.front();
            q.pop();
            if (nn->key_ele == ele) {
                num = num + nn->child.size();
                return num;
            }
            for (int i = 0;
                 i < nn->child.size(); i++) {
                q.push(nn->child[i]);
            }
            n--;
        }
    }
    return num;
}
 
// Function to print the nodes
// having children in range [0, n]
void printTree(Node* root, vector<int> vec)
{
    // map to store nodes
    // with same number of children
    map<int, vector<int> > mp;
    int i = 0;
    for (i = 0; i < vec.size(); i++) {
        int temp;
        temp = numberOfChild(root, vec[i]);
        mp[temp].push_back(vec[i]);
    }
    map<string, int>::iterator iter;
    for (auto& value : mp) {
        cout << value.first << " child: ";
        auto& list = value.second;
        int len = list.size();
        int s = 0;
        for (auto& element : list) {
            if (s < len - 1) {
                cout << element << ", ";
            }
            else {
                cout << element;
            }
            s++;
        }
        cout << endl;
    }
}
 
// Driver Code
int main()
{
    vector<int> vec = { 1, 2, 3, 4, 5, 6, 7 };
    map<int, vector<int> > mp;
    vector<int> v;
    Node* root = new Node(1);
    (root->child).push_back(new Node(2));
    (root->child).push_back(new Node(3));
    (root->child).push_back(new Node(4));
    (root->child[0]->child).push_back(new Node(5));
    (root->child[0]->child).push_back(new Node(6));
    (root->child[2]->child).push_back(new Node(7));
 
    // Function call
    printTree(root, vec);
 
    return 0;
}


Java




// Java code for the above approach:
import java.util.*;
 
public class Main {
    static class Node {
        int key_ele;
        ArrayList <Node> child;
 
        Node(int data_value)
        {
            key_ele = data_value;
            child = new ArrayList <Node> ();
        }
    }
    // Function to find number of child
    static int numberOfChild(Node root, int ele)
    {
        int num = 0;
 
        if (root == null) {
            return 0;
        }
 
        // Initializing queue data structure
        Queue <Node> q = new LinkedList <> ();
        q.add(root);
 
        while (!q.isEmpty()) {
            int n = q.size();
            while (n > 0) {
                Node nn = q.peek();
                q.remove();
                if (nn.key_ele == ele) {
                    num = num + nn.child.size();
                    return num;
                }
                for (int i = 0;
                    i < nn.child.size(); i++) {
                    q.add(nn.child.get(i));
                }
                n--;
            }
        }
        return num;
    }
 
    // Function to print the nodes
    // having children in range [0, n]
    static void printTree(Node root, int[] vec)
    {
        // HashMap to store nodes
        // with same number of children
        HashMap <Integer, ArrayList <Integer> > mp
                            = new HashMap <Integer,ArrayList <Integer>> ();
        int i = 0;
        for (i = 0; i < vec.length; i++) {
            int temp;
            temp = numberOfChild(root, vec[i]);
            if(mp.containsKey(temp)){
                mp.get(temp).add(vec[i]);
            }
            else{
                mp.put(temp, new ArrayList <Integer> ());
                mp.get(temp).add(vec[i]);
            }
        }
 
        for (Map.Entry <Integer, ArrayList<Integer> > value : mp.entrySet()) {
            System.out.print(value.getKey() + " child: ");
            ArrayList <Integer> list = value.getValue();
            int len = list.size();
            for (i=0; i < len; i++) {
                if (i < len - 1) {
                    System.out.print(list.get(i) + ", ");
                }
                else {
                    System.out.print(list.get(i));
                }
            }
            System.out.print("\n");
        }
    }
    public static void main(String args[]) {
        int[] vec = { 1, 2, 3, 4, 5, 6, 7 };
        HashMap <Integer, ArrayList <Integer> > mp;
        ArrayList <Integer> v;
        Node root = new Node(1);
        (root.child).add(new Node(2));
        (root.child).add(new Node(3));
        (root.child).add(new Node(4));
        (root.child.get(0).child).add(new Node(5));
        (root.child.get(0).child).add(new Node(6));
        (root.child.get(2).child).add(new Node(7));
 
        // Function call
        printTree(root, vec);
    }
}
 
// This code has been contributed by Sachin Sahara (sachin801)


Python3




# Node Class
from collections import OrderedDict
class Node:
    def __init__(self, data_value):
        self.key_ele = data_value
        self.child = []
 
# Function to find number of child
def numberOfChild(root, ele):
    num = 0
 
    if root is None:
        return 0
 
    # Initializing queue data structure
    q = []
    q.append(root)
 
    while len(q) > 0:
        n = len(q)
        while n > 0:
            nn = q.pop(0)
            if nn.key_ele == ele:
                num += len(nn.child)
                return num
            for i in range(len(nn.child)):
                q.append(nn.child[i])
            n -= 1
    return num
 
# Function to print the nodes
# having children in range [0, n]
def printTree(root, vec):
   
    # map to store nodes
    # with same number of children
    mp = {}
    for i in range(len(vec)):
        temp = numberOfChild(root, vec[i])
        if temp not in mp:
            mp[temp] = []
        mp[temp].append(vec[i])
    mp = OrderedDict(sorted(mp.items()))
    for key,value in mp.items():
        print(f"{key} child: ", end="")
        len_list = len(value)
        s = 0
        for element in value:
            if s < len_list - 1:
                print(element, end=", ")
            else:
                print(element, end="")
            s += 1
        print()
 
# Driver Code
vec = [1, 2, 3, 4, 5, 6, 7]
mp = {}
v = []
root = Node(1)
root.child.append(Node(2))
root.child.append(Node(3))
root.child.append(Node(4))
root.child[0].child.append(Node(5))
root.child[0].child.append(Node(6))
root.child[2].child.append(Node(7))
 
    # Function call
printTree(root, vec)
 
# This code is contributed by Potta Lokesh


C#




using System;
using System.Collections.Generic;
 
class MainClass {
  class Node {
    public int key_ele;
    public List<Node> child;
 
    public Node(int data_value)
    {
      key_ele = data_value;
      child = new List<Node>();
    }
  }
 
  static int numberOfChild(Node root, int ele)
  {
    int num = 0;
 
    if (root == null) {
      return 0;
    }
 
    Queue<Node> q = new Queue<Node>();
    q.Enqueue(root);
 
    while (q.Count != 0) {
      int n = q.Count;
      while (n > 0) {
        Node nn = q.Peek();
        q.Dequeue();
        if (nn.key_ele == ele) {
          num = num + nn.child.Count;
          return num;
        }
        for (int i = 0; i < nn.child.Count; i++) {
          q.Enqueue(nn.child[i]);
        }
        n--;
      }
    }
    return num;
  }
 
  static void printTree(Node root, int[] vec)
  {
    Dictionary<int, List<int> > mp
      = new Dictionary<int, List<int> >();
    int i = 0;
    for (i = 0; i < vec.Length; i++) {
      int temp = numberOfChild(root, vec[i]);
      if (mp.ContainsKey(temp)) {
        mp[temp].Add(vec[i]);
      }
      else {
        mp.Add(temp, new List<int>());
        mp[temp].Add(vec[i]);
      }
    }
 
    foreach(var value in mp)
    {
      Console.Write(value.Key + " child: ");
      List<int> list = value.Value;
      int len = list.Count;
      for (i = 0; i < len; i++) {
        if (i < len - 1) {
          Console.Write(list[i] + ", ");
        }
        else {
          Console.Write(list[i]);
        }
      }
      Console.WriteLine();
    }
  }
 
  public static void Main(string[] args)
  {
    int[] vec = { 1, 2, 3, 4, 5, 6, 7 };
    Node root = new Node(1);
    root.child.Add(new Node(2));
    root.child.Add(new Node(3));
    root.child.Add(new Node(4));
    root.child[0].child.Add(new Node(5));
    root.child[0].child.Add(new Node(6));
    root.child[2].child.Add(new Node(7));
 
    printTree(root, vec);
  }
}


Javascript




class Node {
    constructor(data_value) {
      this.key_ele = data_value;
      this.child = [];
    }
  }
   
  // Function to find number of child
  function numberOfChild(root, ele) {
    let num = 0;
   
    if (root === null) {
      return 0;
    }
   
    // Initializing queue data structure
    let q = [];
    q.push(root);
   
    while (q.length > 0) {
      let n = q.length;
      while (n > 0) {
        let nn = q.shift();
        if (nn.key_ele === ele) {
          num = num + nn.child.length;
          return num;
        }
        for (let i = 0; i < nn.child.length; i++) {
          q.push(nn.child[i]);
        }
        n--;
      }
    }
    return num;
  }
   
  // Function to print the nodes
  // having children in range [0, n]
  function printTree(root, vec) {
    // map to store nodes
    // with same number of children
    let mp = new Map();
    for (let i = 0; i < vec.length; i++) {
      let temp = numberOfChild(root, vec[i]);
      if (!mp.has(temp)) {
        mp.set(temp, []);
      }
      mp.get(temp).push(vec[i]);
    }
    for (let value of mp) {
      console.log(value[0], "child: ", value[1].join(", "));
    }
  }
   
  // Driver Code
  let vec = [1, 2, 3, 4, 5, 6, 7];
  let root = new Node(1);
  root.child.push(new Node(2));
  root.child.push(new Node(3));
  root.child.push(new Node(4));
  root.child[0].child.push(new Node(5));
  root.child[0].child.push(new Node(6));
  root.child[2].child.push(new Node(7));
   
  // Function call
  printTree(root, vec);
   
  // This code is contributed by aadityamaharshi21.


Output

0 child: 3, 5, 6, 7
1 child: 4
2 child: 2
3 child: 1

Time Complexity: O(N2)
Auxiliary Space: O(N)


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