Print levels with odd number of nodes and even number of nodes

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 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:

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// 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;
}

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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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Striver(underscore)79 at Codechef and codeforces D

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