Print all the levels with odd and even number of nodes in it

Print all the levels with odd and even number of nodes in it | Set-2

Last Updated : 27 Oct, 2020

Given an N-ary tree, print all the levels with odd and even number 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 BFS on the tree such that height[node] = 1 + height[parent]
Once BFS 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++

// C++ program to print all levels
// with odd and even number of nodes
 
#include <bits/stdc++.h>
using namespace std;
 
// Function for BFS in a tree
void bfs(int node, int parent, int height[], int vis[],
         vector<int> tree[])
{
 
    // mark first node as visited
    vis[node] = 1;
 
    // Declare Queue
    queue<int> q;
 
    // Push the first element
    q.push(1);
 
    // calculate the level of every node
    height[node] = 1 + height[parent];
 
    // Check if the queue is empty or not
    while (!q.empty()) {
 
        // Get the top element in the queue
        int top = q.front();
 
        // pop the element
        q.pop();
 
        // mark as visited
        vis[top] = 1;
 
        // Iterate for the connected nodes
        for (int i = 0; i < tree[top].size(); i++) {
 
            // if not visited
            if (!vis[tree[top][i]]) {
 
                // Insert into queue
                q.push(tree[top][i]);
 
                // Increase level
                height[tree[top][i]] = 1 + height[top];
            }
        }
    }
}
 
// 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 };
 
    // call the bfs function
    bfs(1, 0, height, vis, tree);
 
    // Function to print
    printLevelsOddEven(N, vis, height);
 
    return 0;
}

Python3

# Python3 program to print all levels 
# with odd and even number of nodes 
 
# Function for BFS in a tree 
def bfs(node, parent, height, vis, tree):
 
    # mark first node as visited 
    vis[node] = 1
 
    # Declare Queue 
    q = [] 
 
    # append the first element 
    q.append(1) 
 
    # calculate the level of every node 
    height[node] = 1 + height[parent] 
 
    # Check if the queue is empty or not 
    while (len(q)):
 
        # Get the top element in 
        # the queue 
        top = q[0] 
 
        # pop the element 
        q.pop(0) 
 
        # mark as visited 
        vis[top] = 1
 
        # Iterate for the connected nodes 
        for i in range(len(tree[top])): 
             
            # if not visited 
            if (not vis[tree[top][i]]): 
 
                # Insert into queue 
                q.append(tree[top][i]) 
 
                # Increase level 
                height[tree[top][i]] = 1 + height[top] 
 
# 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) 
     
    # prodd 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 ): 
        if (mark[i] % 2 == 0): 
            print(i, end = " ")
 
# Driver Code 
if __name__ == '__main__':
     
    # Construct the tree 
    """ 1 
    / \ 
    2 3 
    / \ \ 
    4 5 6 
        / \ / 
    7 8 9 """
 
    N = 9
 
    tree = [[0]] * (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 bfs function 
    bfs(1, 0, height, vis, tree) 
 
    # Function to pr
    printLevelsOddEven(N, vis, height)
 
# This code is contributed 
# by SHUBHAMSINGH10

C#

// C# program to print all levels
// with odd and even number of nodes
using System;
using System.Collections;
 
class GFG{
 
// Function for BFS in a tree
static void bfs(int node, int parent, 
                int []height, int []vis,
                ArrayList []tree)
{
     
    // Mark first node as visited
    vis[node] = 1;
  
    // Declare Queue
    Queue q = new Queue();
  
    // Push the first element
    q.Enqueue(1);
  
    // Calculate the level of every node
    height[node] = 1 + height[parent];
  
    // Check if the queue is empty or not
    while (q.Count != 0)
    {
         
        // Get the top element in the queue
        int top = (int)q.Peek();
  
        // Pop the element
        q.Dequeue();
  
        // Mark as visited
        vis[top] = 1;
  
        // Iterate for the connected nodes
        for(int i = 0; i < tree[top].Count; i++) 
        {
             
            // If not visited
            if (vis[(int)tree[top][i]] == 0)
            {
                 
                // Insert into queue
                q.Enqueue(tree[top][i]);
  
                // Increase level
                height[(int)tree[top][i]] = 1 + height[top];
            }
        }
    }
}
  
// 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 bfs function
    bfs(1, 0, height, vis, tree);
     
    // Function to print
    printLevelsOddEven(N, vis, height);
}
}
 
// This code is contributed by rutvik_56

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)
Auxiliary Space: O(N)

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