Level Order Traversal of N-ary Tree
Last Updated :
17 Mar, 2023
Given an N-ary Tree. The task is to print the level order traversal of the tree where each level will be in a new line.
Examples:
Input:
Image
Output:
1
3 2 4
5 6
Explanation: At level 1: only 1 is present.
At level 2: 3, 2, 4 is present
At level 3: 5, 6 is present
Input:
Image
Output:
1
2 3 4 5
6 7 8 9 10
11 12 13
14
Explanation: For the above example there are 5 level present in the n-ary tree.
At level 1: only 1 is present.
At level 2: 2, 3, 4, 5 is present.
At level 3: 6, 7, 8, 9, 10 is present
At level 4:11, 12, 13 is present
At level 5 :- 14 is present
Approach 1: Using BFS
The approach of the problem is to use Level Order Traversal and store all the levels in a 2D array where each of the levels is stored in a different row.
Follow the below steps to implement the approach:
- Create a vector ans and temp to store the level order traversal of the N-ary tree.
- Push the root node in the queue.
- Run a while loop until the queue is not empty:
- Determine the size of the current level which is the size of the queue (say N):
- Run a loop for i = 1 to N
- In each step delete the front node (say cur) and push its data to the temp as a part of the current level.
- Push all the children of cur into the queue.
- Push the temp into the final ans vector which stores the different levels in different rows.
- Return the ans vector.
Below is the implementation of the above approach:
C++
// C++ code for above implementation
#include <bits/stdc++.h>
using namespace std;
struct Node {
char val;
vector<Node*> children;
};
// Utility function to create a new tree node
Node* newNode(int key)
{
Node* temp = new Node;
temp->val = key;
return temp;
}
// Function for level order traversal for n-array tree
vector<vector<int> > levelOrder(Node* root)
{
vector<vector<int> > ans;
if (!root)
cout << "N-Ary tree does not any nodes";
// Create a queue namely main_queue
queue<Node*> main_queue;
// Push the root value in the main_queue
main_queue.push(root);
// Create a temp vector to store the all the node values
// present at a particular level
vector<int> temp;
// Run a while loop until the main_queue is empty
while (!main_queue.empty()) {
// Get the front of the main_queue
int n = main_queue.size();
// Iterate through the current level
for (int i = 0; i < n; i++) {
Node* cur = main_queue.front();
main_queue.pop();
temp.push_back(cur->val);
for (auto u : cur->children)
main_queue.push(u);
}
ans.push_back(temp);
temp.clear();
}
return ans;
}
// Driver code
int main()
{
Node* root = newNode(1);
root->children.push_back(newNode(3));
root->children.push_back(newNode(2));
root->children.push_back(newNode(4));
root->children[0]->children.push_back(newNode(5));
root->children[0]->children.push_back(newNode(6));
// LevelOrderTraversal obj;
vector<vector<int> > ans = levelOrder(root);
for (auto v : ans) {
for (int x : v)
cout << x << " ";
cout << endl;
}
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
import java.util.*;
public class Main {
static class Node {
public int val;
public Vector<Node> children;
public Node(int key)
{
val = key;
children = new Vector<Node>();
}
}
// Utility function to create a new tree node
static Node newNode(int key)
{
Node temp = new Node(key);
return temp;
}
// Function for level order traversal for n-array tree
static List<List<Integer> > levelOrder(Node root)
{
List<List<Integer> > ans = new ArrayList<>();
if (root == null)
System.out.println(
"N-Ary tree does not any nodes");
// Create one queue main_queue
Queue<Node> main_queue = new LinkedList<>();
// Push the root value in the main_queue
main_queue.offer(root);
// Traverse the N-ary Tree by level
while (!main_queue.isEmpty()) {
// Create a temp vector to store the all the
// node values present at a particular level
List<Integer> temp = new ArrayList<>();
int size = main_queue.size();
// Iterate through the current level
for (int i = 0; i < size; i++) {
Node node = main_queue.poll();
temp.add(node.val);
for (Node child : node.children) {
main_queue.offer(child);
}
}
ans.add(temp);
}
return ans;
}
// Utility function to print the level order traversal
public static void printList(List<List<Integer> > temp)
{
for (List<Integer> it : temp) {
for (Integer et : it)
System.out.print(et + " ");
System.out.println();
}
}
public static void main(String[] args)
{
Node root = newNode(1);
(root.children).add(newNode(3));
(root.children).add(newNode(2));
(root.children).add(newNode(4));
(root.children.get(0).children).add(newNode(5));
(root.children.get(0).children).add(newNode(6));
List<List<Integer> > ans = levelOrder(root);
printList(ans);
}
}
// This code is contributed by Sania Kumari Gupta
Python3
# Python code for above implementation
class Node:
def __init__(self, key):
self.key = key
self.child = []
# Utility function to create a new tree node
def newNode(key):
temp = Node(key)
return temp
# Prints the n-ary tree level wise
def LevelOrderTraversal(root):
if (root == None):
return;
# Standard level order traversal code
# using queue
q = [] # Create a queue
q.append(root); # Enqueue root
while (len(q) != 0):
n = len(q);
# If this node has children
while (n > 0):
# Dequeue an item from queue and print it
p = q[0]
q.pop(0);
print(p.key, end=' ')
# Enqueue all children of the dequeued item
for i in range(len(p.child)):
q.append(p.child[i]);
n -= 1
print() # Print new line between two levels
if __name__ == '__main__':
root = newNode(1);
root.child.append(newNode(3));
root.child.append(newNode(2));
root.child.append(newNode(4));
root.child[0].child.append(newNode(5));
root.child[0].child.append(newNode(6));
# LevelOrderTraversal obj;
LevelOrderTraversal(root);
# This code is contributed by poojaagarwal2.
C#
// C# code implementation for the abvoe approach
using System;
using System.Collections.Generic;
using System.Linq;
public class Node {
public int val;
public List<Node> children;
public Node(int key)
{
val = key;
children = new List<Node>();
}
}
public class GFG {
// Utility function to create a new tree node
static Node newNode(int key)
{
Node temp = new Node(key);
return temp;
}
// Function for level order traversal for n-array tree
static List<List<int> > levelOrder(Node root)
{
List<List<int> > ans = new List<List<int> >();
if (root == null)
Console.WriteLine(
"N-Ary tree does not any nodes");
// Create one queue main_queue
Queue<Node> main_queue = new Queue<Node>();
// Push the root value in the main_queue
main_queue.Enqueue(root);
// Traverse the N-ary Tree by level
while (main_queue.Any()) {
// Create a temp vector to store the all the
// node values present at a particular level
List<int> temp = new List<int>();
int size = main_queue.Count;
// Iterate through the current level
for (int i = 0; i < size; i++) {
Node node = main_queue.Dequeue();
temp.Add(node.val);
foreach(Node child in node.children)
{
main_queue.Enqueue(child);
}
}
ans.Add(temp);
}
return ans;
}
// Utility function to print the level order traversal
public static void printList(List<List<int> > temp)
{
foreach(List<int> it in temp)
{
foreach(int et in it) Console.Write(et + " ");
Console.WriteLine();
}
}
static public void Main()
{
// Code
Node root = newNode(1);
(root.children).Add(newNode(3));
(root.children).Add(newNode(2));
(root.children).Add(newNode(4));
(root.children[0].children).Add(newNode(5));
(root.children[0].children).Add(newNode(6));
List<List<int> > ans = levelOrder(root);
printList(ans);
}
}
// This code is contributed by karthik.
JavaScript
// Javascript code for above implementation
class Node {
constructor(val) {
this.val = val;
this.children = new Array();
}
}
// Function for level order traversal for n-array tree
function levelOrder( root)
{
let ans = [];
if (!root)
console.log("N-Ary tree does not any nodes");
// Create a queue namely main_queue
let main_queue=[];
// Push the root value in the main_queue
main_queue.push(root);
// Create a temp vector to store the all the node values
// present at a particular level
let temp=[];
// Run a while loop until the main_queue is empty
while (main_queue.length) {
// Get the front of the main_queue
let n = main_queue.length;
// Iterate through the current level
for (let i = 0; i < n; i++) {
let cur = main_queue.shift();
temp.push(cur.val);
for (let u of cur.children)
main_queue.push(u);
}
ans.push(temp);
temp=[];
}
return ans;
}
// Driver code
let root = new Node(1);
root.children.push(new Node(3));
root.children.push(new Node(2));
root.children.push(new Node(4));
root.children[0].children.push(new Node(5));
root.children[0].children.push(new Node(6));
// LevelOrderTraversal obj;
let ans = levelOrder(root);
for (let v of ans) {
for (let x of v)
console.log(x+" ");
console.log("<br>");
}
Time Complexity: O(V) where V is the number of nodes
Auxiliary Space: O(V)
Approach 2: Using DFS
The approach of the problem is to use Level Order Traversal using DFS and store all the levels in a 2D array where each of the levels is stored in a different row.
- LevelOrder function will update ans with the current value, pushing it in with a new sub-vector if one matching the level is not present already into ans.
- Function will increase level by 1;
- It will call itself recursively on all the children;
- It will backtrack level.
Below is the implementation of the above approach:
C++
// C++ code for above implementation
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > ans;
int level = 0;
struct Node {
char val;
vector<Node*> children;
};
Node* newNode(int key)
{
Node* temp = new Node;
temp->val = key;
return temp;
}
void levelOrder(Node *root) {
if (ans.size() == level) ans.push_back({root->val});
else ans[level].push_back(root->val);
level++;
for (Node *n: root->children) levelOrder(n);
level--;
}
int main()
{
Node* root = newNode(1);
root->children.push_back(newNode(3));
root->children.push_back(newNode(2));
root->children.push_back(newNode(4));
root->children[0]->children.push_back(newNode(5));
root->children[0]->children.push_back(newNode(6));
// LevelOrderTraversal obj;
levelOrder(root);
for (auto v : ans) {
for (int x : v)
cout << x << " ";
cout << endl;
}
return 0;
}
Java
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
public class GFG {
static List<List<Integer> > result = new ArrayList<>();
static int level = 0;
static class Node {
public int val;
public Vector<Node> children;
public Node(int key)
{
val = key;
children = new Vector<Node>();
}
}
// Utility function to create a new tree node
static Node newNode(int key)
{
Node temp = new Node(key);
return temp;
}
// method to find level order traversal of n-ary tree
static void levelOrder(Node node)
{
if (node == null) {
return;
}
List<Integer> list = result.size() > level
? result.get(level)
: new ArrayList<>();
// adding node value to the list
list.add(node.val);
if (result.size() <= level) {
result.add(list);
}
// promoting/incrementing the level to next
level++;
for (Node n : node.children) {
levelOrder(n);
}
level--;
}
// utility function to print level order traversal
public static void printList(List<List<Integer> > temp)
{
for (List<Integer> it : temp) {
for (Integer et : it)
System.out.print(et + " ");
System.out.println();
}
}
public static void main(String[] args)
{
Node root = newNode(1);
(root.children).add(newNode(3));
(root.children).add(newNode(2));
(root.children).add(newNode(4));
(root.children.get(0).children).add(newNode(5));
(root.children.get(0).children).add(newNode(6));
levelOrder(root);
printList(result);
}
}
Python3
# Python code for above implementation
ans = []
level = 0
class Node:
def __init__(self, val):
self.val = val
self.children = []
def levelOrder(root):
global level
if len(ans) == level:
ans.append([root.val])
else:
ans[level].append(root.val)
level += 1
for n in root.children:
levelOrder(n)
level -= 1
root = Node(1)
root.children.append(Node(3))
root.children.append(Node(2))
root.children.append(Node(4))
root.children[0].children.append(Node(5))
root.children[0].children.append(Node(6))
levelOrder(root)
for v in ans:
for x in v:
print(x, end=" ")
print()
# This code is contributed by Tapesh(tapeshdua420)
C#
// C# code for the above approach
using System;
using System.Collections.Generic;
public class GFG {
static List<List<int> > result = new List<List<int> >();
static int level = 0;
public class Node {
public int val;
public List<Node> children;
public Node(int key)
{
val = key;
children = new List<Node>();
}
}
// Utility function to create a new tree node
static Node newNode(int key)
{
Node temp = new Node(key);
return temp;
}
// method to find level order traversal of n-ary tree
static void levelOrder(Node node)
{
if (node == null) {
return;
}
List<int> list = result.Count > level
? result[level]
: new List<int>();
// adding node value to the list
list.Add(node.val);
if (result.Count <= level) {
result.Add(list);
}
// promoting/incrementing the level to next
level++;
foreach(Node n in node.children) { levelOrder(n); }
level--;
}
// utility function to print level order traversal
public static void printList(List<List<int> > temp)
{
foreach(List<int> it in temp)
{
foreach(int et in it)
{
Console.Write(et + " ");
}
Console.WriteLine();
}
}
static public void Main()
{
// Code
Node root = newNode(1);
(root.children).Add(newNode(3));
(root.children).Add(newNode(2));
(root.children).Add(newNode(4));
(root.children[0].children).Add(newNode(5));
(root.children[0].children).Add(newNode(6));
levelOrder(root);
printList(result);
}
}
// This code is contributed by sankar.
JavaScript
// JavaScript code for the above approach
const ans = [];
let level = 0;
class Node {
constructor(val) {
this.val = val;
this.children = [];
}
}
function levelOrder(root) {
if (ans.length === level) ans.push([root.val]);
else ans[level].push(root.val);
level++;
for (const n of root.children) levelOrder(n);
level--;
}
// create tree
const root = new Node(1);
root.children.push(new Node(3));
root.children.push(new Node(2));
root.children.push(new Node(4));
root.children[0].children.push(new Node(5));
root.children[0].children.push(new Node(6));
levelOrder(root);
for (const v of ans) {
for (const x of v) {
console.log(x + " ");
}
console.log('<br>');
}
// This code is contributed by Potta Lokesh
Time Complexity: O(V)
Auxiliary Space: O(V)
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Minimum distance between two given nodes in an N-ary treeGiven a N ary Tree consisting of N nodes, the task is to find the minimum distance from node A to node B of the tree. Examples: Input: 1 / \ 2 3 / \ / \ \4 5 6 7 8A = 4, B = 3Output: 3Explanation: The path 4->2->1->3 gives the minimum distance from A to B. Input: 1 / \ 2 3 / \ \ 6 7 8A = 6,
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Average width in a N-ary treeGiven a Generic Tree consisting of N nodes, the task is to find the average width for each node present in the given tree. The average width for each node can be calculated by the ratio of the total number of nodes in that subtree(including the node itself) to the total number of levels under that n
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Maximum width of an N-ary treeGiven an N-ary tree, the task is to find the maximum width of the given tree. The maximum width of a tree is the maximum of width among all levels. Examples: Input: 4 / | \ 2 3 -5 / \ /\ -1 3 -2 6 Output: 4 Explanation: Width of 0th level is 1. Width of 1st level is 3. Width of 2nd level is 4. There
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