Given an N-ary tree, print all the levels with odd and even numbers of nodes in it.
Examples:
For example consider the following tree
1 - Level 1
/ \
2 3 - Level 2
/ \ \
4 5 6 - Level 3
/ \ /
7 8 9 - Level 4
The levels with odd number of nodes are: 1 3 4
The levels with even number of nodes are: 2
Note: The level numbers starts from 1. That is, the root node is at the level 1.
Approach:
- Insert all the connecting nodes to a 2-D vector tree.
- Run a DFS on the tree such that height[node] = 1 + height[parent]
- Once DFS traversal is completed, increase the count[] array by 1, for every node's level.
- Iterate from first level to last level, and print all nodes with count[] values as odd to get level with odd number nodes.
- Iterate from first level to last level, and print all nodes with count[] values as even to get level with even number nodes.
Below is the implementation of the above approach:
// C++ program to print all levels
// with odd and even number of nodes
#include <bits/stdc++.h>
using namespace std;
// Function for DFS in a tree
void dfs(int node, int parent, int height[], int vis[],
vector<int> tree[])
{
// calculate the level of every node
height[node] = 1 + height[parent];
// mark every node as visited
vis[node] = 1;
// iterate in the subtree
for (auto it : tree[node]) {
// if the node is not visited
if (!vis[it]) {
// call the dfs function
dfs(it, node, height, vis, tree);
}
}
}
// Function to insert edges
void insertEdges(int x, int y, vector<int> tree[])
{
tree[x].push_back(y);
tree[y].push_back(x);
}
// Function to print all levels
void printLevelsOddEven(int N, int vis[], int height[])
{
int mark[N + 1];
memset(mark, 0, sizeof mark);
int maxLevel = 0;
for (int i = 1; i <= N; i++) {
// count number of nodes
// in every level
if (vis[i])
mark[height[i]]++;
// find the maximum height of tree
maxLevel = max(height[i], maxLevel);
}
// print odd number of nodes
cout << "The levels with odd number of nodes are: ";
for (int i = 1; i <= maxLevel; i++) {
if (mark[i] % 2)
cout << i << " ";
}
// print even number of nodes
cout << "\nThe levels with even number of nodes are: ";
for (int i = 1; i <= maxLevel; i++) {
if (mark[i] % 2 == 0)
cout << i << " ";
}
}
// Driver Code
int main()
{
// Construct the tree
/* 1
/ \
2 3
/ \ \
4 5 6
/ \ /
7 8 9 */
const int N = 9;
vector<int> tree[N + 1];
insertEdges(1, 2, tree);
insertEdges(1, 3, tree);
insertEdges(2, 4, tree);
insertEdges(2, 5, tree);
insertEdges(5, 7, tree);
insertEdges(5, 8, tree);
insertEdges(3, 6, tree);
insertEdges(6, 9, tree);
int height[N + 1];
int vis[N + 1] = { 0 };
height[0] = 0;
// call the dfs function
dfs(1, 0, height, vis, tree);
// Function to print
printLevelsOddEven(N, vis, height);
return 0;
}
// Java program to print all levels
// with odd and even number of nodes
import java.util.*;
@SuppressWarnings("unchecked")
class GFG{
// Function for DFS in a tree
static void dfs(int node, int parent,
int []height, int []vis,
ArrayList []tree)
{
// Calculate the level of every node
height[node] = 1 + height[parent];
// Mark every node as visited
vis[node] = 1;
// Iterate in the subtree
for(int it : (ArrayList<Integer>)tree[node])
{
// If the node is not visited
if (vis[it] == 0)
{
// Call the dfs function
dfs(it, node, height, vis, tree);
}
}
}
// Function to insert edges
static void insertEdges(int x, int y,
ArrayList []tree)
{
tree[x].add(y);
tree[y].add(x);
}
// Function to print all levels
static void printLevelsOddEven(int N, int []vis,
int []height)
{
int []mark = new int[N + 1];
Arrays.fill(mark, 0);
int maxLevel = 0;
for(int i = 1; i <= N; i++)
{
// Count number of nodes
// in every level
if (vis[i] != 0)
mark[height[i]]++;
// Find the maximum height of tree
maxLevel = Math.max(height[i], maxLevel);
}
// Print odd number of nodes
System.out.print("The levels with odd " +
"number of nodes are: ");
for(int i = 1; i <= maxLevel; i++)
{
if (mark[i] % 2 != 0)
{
System.out.print(i + " ");
}
}
// Print even number of nodes
System.out.print("\nThe levels with even " +
"number of nodes are: ");
for(int i = 1; i <= maxLevel; i++)
{
if (mark[i] % 2 == 0)
{
System.out.print(i + " ");
}
}
}
// Driver code
public static void main(String []s)
{
// Construct the tree
/* 1
/ \
2 3
/ \ \
4 5 6
/ \ /
7 8 9 */
int N = 9;
ArrayList []tree = new ArrayList[N + 1];
for(int i = 0; i < N + 1; i++)
{
tree[i] = new ArrayList();
}
insertEdges(1, 2, tree);
insertEdges(1, 3, tree);
insertEdges(2, 4, tree);
insertEdges(2, 5, tree);
insertEdges(5, 7, tree);
insertEdges(5, 8, tree);
insertEdges(3, 6, tree);
insertEdges(6, 9, tree);
int []height = new int[N + 1];
int []vis = new int[N + 1];
Arrays.fill(vis, 0);
height[0] = 0;
// Call the dfs function
dfs(1, 0, height, vis, tree);
// Function to print
printLevelsOddEven(N, vis, height);
}
}
// This code is contributed by pratham76
// C# program to print all levels
// with odd and even number of nodes
using System;
using System.Collections;
class GFG{
// Function for DFS in a tree
static void dfs(int node, int parent,
int []height, int []vis,
ArrayList []tree)
{
// Calculate the level of every node
height[node] = 1 + height[parent];
// Mark every node as visited
vis[node] = 1;
// Iterate in the subtree
foreach (int it in tree[node])
{
// If the node is not visited
if (vis[it] == 0)
{
// Call the dfs function
dfs(it, node, height, vis, tree);
}
}
}
// Function to insert edges
static void insertEdges(int x, int y,
ArrayList []tree)
{
tree[x].Add(y);
tree[y].Add(x);
}
// Function to print all levels
static void printLevelsOddEven(int N, int []vis,
int []height)
{
int []mark = new int[N + 1];
Array.Fill(mark, 0);
int maxLevel = 0;
for(int i = 1; i <= N; i++)
{
// Count number of nodes
// in every level
if (vis[i] != 0)
mark[height[i]]++;
// Find the maximum height of tree
maxLevel = Math.Max(height[i], maxLevel);
}
// Print odd number of nodes
Console.Write("The levels with odd " +
"number of nodes are: ");
for(int i = 1; i <= maxLevel; i++)
{
if (mark[i] % 2 != 0)
{
Console.Write(i + " ");
}
}
// Print even number of nodes
Console.Write("\nThe levels with even " +
"number of nodes are: ");
for(int i = 1; i <= maxLevel; i++)
{
if (mark[i] % 2 == 0)
{
Console.Write(i + " ");
}
}
}
// Driver code
static void Main()
{
// Construct the tree
/* 1
/ \
2 3
/ \ \
4 5 6
/ \ /
7 8 9 */
int N = 9;
ArrayList []tree = new ArrayList[N + 1];
for(int i = 0; i < N + 1; i++)
{
tree[i] = new ArrayList();
}
insertEdges(1, 2, tree);
insertEdges(1, 3, tree);
insertEdges(2, 4, tree);
insertEdges(2, 5, tree);
insertEdges(5, 7, tree);
insertEdges(5, 8, tree);
insertEdges(3, 6, tree);
insertEdges(6, 9, tree);
int []height = new int[N + 1];
int []vis = new int[N + 1];
Array.Fill(vis, 0);
height[0] = 0;
// Call the dfs function
dfs(1, 0, height, vis, tree);
// Function to print
printLevelsOddEven(N, vis, height);
}
}
// This code is contributed by rutvik_56
// JavaScript program to print all levels
// with odd and even number of nodes
// Function for DFS in a tree
function dfs(node, parent, height, vis, tree) {
// Calculate the level of every node
height[node] = 1 + height[parent];
// Mark every node as visited
vis[node] = 1;
// Iterate in the subtree
for (let it = 0; it < tree[node].length; it++) {
// If the node is not visited
if (vis[tree[node][it]] == 0) {
// Call the dfs function
dfs(tree[node][it], node, height, vis, tree);
}
}
}
// Function to insert edges
function insertEdges(x, y, tree) {
tree[x].push(y);
tree[y].push(x);
}
// Function to print all levels
function printLevelsOddEven(N, vis, height) {
let mark = new Array(N + 1);
mark.fill(0);
let maxLevel = 0;
for (let i = 1; i <= N; i++) {
// Count number of nodes
// in every level
if (vis[i] != 0)
mark[height[i]]++;
// Find the maximum height of tree
maxLevel = Math.max(height[i], maxLevel);
}
// Print odd number of nodes
console.log("The levels with odd " +
"number of nodes are: ");
for (let i = 1; i <= maxLevel; i++) {
if (mark[i] % 2 != 0) {
console.log(i + " ");
}
}
// Print even number of nodes
console.log("The levels with even " +
"number of nodes are: ");
for (let i = 1; i <= maxLevel; i++) {
if (mark[i] % 2 == 0) {
console.log(i + " ");
}
}
}
// Construct the tree
/* 1
/ \
2 3
/ \ \
4 5 6
/ \ /
7 8 9 */
let N = 9;
let tree = new Array(N + 1);
for (let i = 0; i < N + 1; i++) {
tree[i] = [];
}
insertEdges(1, 2, tree);
insertEdges(1, 3, tree);
insertEdges(2, 4, tree);
insertEdges(2, 5, tree);
insertEdges(5, 7, tree);
insertEdges(5, 8, tree);
insertEdges(3, 6, tree);
insertEdges(6, 9, tree);
let height = new Array(N + 1);
let vis = new Array(N + 1);
vis.fill(0);
height[0] = 0;
// Call the dfs function
dfs(1, 0, height, vis, tree);
// Function to print
printLevelsOddEven(N, vis, height);
# Python3 program to print all levels
# with odd and even number of nodes
# Function for DFS in a tree
def dfs(node, parent, height, vis, tree):
# calculate the level of every node
height[node] = 1 + height[parent]
# mark every node as visited
vis[node] = 1
# iterate in the subtree
for it in tree[node]:
# if the node is not visited
if not vis[it]:
# call the dfs function
dfs(it, node, height, vis, tree)
# Function to insert edges
def insertEdges(x, y, tree):
tree[x].append(y)
tree[y].append(x)
# Function to print all levels
def printLevelsOddEven(N, vis, height):
mark = [0] * (N + 1)
maxLevel = 0
for i in range(1, N + 1):
# count number of nodes in every level
if vis[i]:
mark[height[i]] += 1
# find the maximum height of tree
maxLevel = max(height[i], maxLevel)
# print odd number of nodes
print("The levels with odd number",
"of nodes are: ", end = "")
for i in range(1, maxLevel + 1):
if mark[i] % 2:
print(i, end = " ")
# print even number of nodes
print("\nThe levels with even number",
"of nodes are: ", end = "")
for i in range(1, maxLevel + 1):
if mark[i] % 2 == 0:
print(i, end = " ")
# Driver Code
if __name__ == "__main__":
# Construct the tree
N = 9
tree = [[] for i in range(N + 1)]
insertEdges(1, 2, tree)
insertEdges(1, 3, tree)
insertEdges(2, 4, tree)
insertEdges(2, 5, tree)
insertEdges(5, 7, tree)
insertEdges(5, 8, tree)
insertEdges(3, 6, tree)
insertEdges(6, 9, tree)
height = [0] * (N + 1)
vis = [0] * (N + 1)
# call the dfs function
dfs(1, 0, height, vis, tree)
# Function to print
printLevelsOddEven(N, vis, height)
# This code is contributed by Rituraj Jain
Output
The levels with odd number of nodes are: 1 3 4 The levels with even number of nodes are: 2
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)
Python Solution(using dequeue):
Approach:
Utilizes a queue (deque from the collections module) to traverse the tree level by level. Each node in the queue is accompanied by its corresponding level.
While traversing the tree, the code counts the number of nodes at each level by incrementing the count in the result list. If the level exceeds the length of the result list, it appends a new entry.
After traversing the tree and counting nodes at each level, it iterates through the counts and identifies levels with odd and even numbers of nodes, storing them in separate lists (odd_levels and even_levels).
from collections import deque
# Function to perform level-order traversal and count nodes at each level
def countNodesAtEachLevel(root):
if not root:
return []
result = []
queue = deque([(root, 1)]) # Tuple contains the node and its level
while queue:
node, level = queue.popleft()
if len(result) < level:
result.append(0)
result[level-1] += 1
for child in node.children:
queue.append((child, level + 1))
return result
# Function to print levels with odd and even numbers of nodes
def printLevelsOddEven(counts):
odd_levels = []
even_levels = []
for level, count in enumerate(counts, start=1):
if count % 2 == 1:
odd_levels.append(level)
else:
even_levels.append(level)
print("The levels with odd number of nodes are:", *odd_levels)
print("The levels with even number of nodes are:", *even_levels)
# Define a class for the tree node
class TreeNode:
def __init__(self, val=0, children=None):
self.val = val
self.children = children if children is not None else []
# Driver code
if __name__ == "__main__":
# Construct the tree
root = TreeNode(1)
root.children = [TreeNode(2), TreeNode(3)]
root.children[0].children = [TreeNode(4), TreeNode(5)]
root.children[1].children = [TreeNode(6)]
root.children[0].children[1].children = [TreeNode(7), TreeNode(8)]
root.children[1].children[0].children = [TreeNode(9)]
# Perform level-order traversal and count nodes at each level
counts = countNodesAtEachLevel(root)
# Print levels with odd and even numbers of nodes
printLevelsOddEven(counts)
Output
The levels with odd number of nodes are: 1 3 4 The levels with even number of nodes are: 2
Time complexity: O(n + h), where n is the number of nodes and h is the height of tree.
Space Complexity: O(h), where h is the height of tree.