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++

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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 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;
}

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Python3

# Python3 program to prall 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 prall 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

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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