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:
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)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.