Print all the levels with odd and even number of nodes in it | Set-2
Given an N-ary tree, print all the levels with odd and even number of nodes in it.
Examples:
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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: 2Note: 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 treevoid 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 edgesvoid insertEdges(int x, int y, vector<int> tree[]){ tree[x].push_back(y); tree[y].push_back(x);}// Function to print all levelsvoid 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 Codeint 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;} |
Java
// Java program to print all levels// with odd and even number of nodesimport java.util.*;public class Main{ // Function for BFS in a tree static void bfs(int node, int parent, int[] height, int[] vis, Vector<Vector<Integer>> tree) { // Mark first node as visited vis[node] = 1; // Declare Queue Queue<Integer> q = new LinkedList<>(); // Push the first element q.add(1); // Calculate the level of every node height[node] = 1 + height[parent]; // Check if the queue is empty or not while (q.size() != 0) { // Get the top element in the queue int top = (int)q.peek(); // Pop the element q.remove(); // Mark as visited vis[top] = 1; // Iterate for the connected nodes for(int i = 0; i < tree.get(top).size(); i++) { // If not visited if (vis[(int)tree.get(top).get(i)] == 0) { // Insert into queue q.add(tree.get(top).get(i)); // Increase level height[(int)tree.get(top).get(i)] = 1 + height[top]; } } } } // Function to insert edges static void insertEdges(int x, int y, Vector<Vector<Integer>> tree) { tree.get(x).add(y); tree.get(y).add(x); } // Function to print all levels static void printLevelsOddEven(int N, int[] vis, int[] height) { int[] mark = new int[N + 1]; for(int i = 0; i < N + 1; i++) { mark[i] = 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.println(); System.out.print("The levels with even " + "number of nodes are: "); for(int i = 1; i <= maxLevel; i++) { if (mark[i] % 2 == 0) { System.out.print(i + " "); } } } public static void main(String[] args) { // Construct the tree /* 1 / \ 2 3 / \ \ 4 5 6 / \ / 7 8 9 */ int N = 9; Vector<Vector<Integer>> tree = new Vector<Vector<Integer>>(); for(int i = 0; i < N + 1; i++) { tree.add(new Vector<Integer>()); } 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]; for(int i = 0; i < N + 1; i++) { vis[i] = 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 divyeshrabadiya07. |
Python3
# Python3 program to print all levels# with odd and even number of nodes# Function for BFS in a treedef 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 edgesdef insertEdges(x, y, tree): tree[x].append(y) tree[y].append(x) # Function to print all levelsdef 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 Codeif __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 nodesusing System;using System.Collections;class GFG{// Function for BFS in a treestatic 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 edgesstatic void insertEdges(int x, int y, ArrayList []tree){ tree[x].Add(y); tree[y].Add(x);} // Function to print all levelsstatic 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 |
Javascript
<script> // JavaScript program to print all levels // with odd and even number of nodes // Function for BFS in a tree function bfs(node, parent, height, vis, tree) { // Mark first node as visited vis[node] = 1; // Declare Queue let 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.length != 0) { // Get the top element in the queue let top = q[0]; // Pop the element q.shift(); // Mark as visited vis[top] = 1; // Iterate for the connected nodes for(let i = 0; i < tree[top].length; i++) { // If not visited if (vis[tree[top][i]] == 0) { // Insert into queue q.push(tree[top][i]); // Increase level height[tree[top][i]] = 1 + height[top]; } } } } // 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 document.write("The levels with odd " + "number of nodes are: "); for(let i = 1; i <= maxLevel; i++) { if (mark[i] % 2 != 0) { document.write(i + " "); } } // print even number of nodes document.write("</br>" + "The levels with even " + "number of nodes are: "); for(let i = 1; i <= maxLevel; i++) { if (mark[i] % 2 == 0) { document.write(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 bfs function bfs(1, 0, height, vis, tree); // Function to print printLevelsOddEven(N, vis, height);</script> |
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)
