Dado un árbol N-ario , imprima todos los niveles con un número par e impar de Nodes.
Ejemplos :
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
Nota : los números de nivel comienzan desde 1. Es decir, el Node raíz está en el nivel 1.
Enfoque :
- Inserte todos los Nodes de conexión en un árbol vectorial 2D.
- Ejecute un BFS en el árbol de modo que altura[Node] = 1 + altura[padre]
- Una vez que se completa el recorrido de BFS, aumente la array count[] en 1, para el nivel de cada Node.
- Iterar desde el primer nivel hasta el último nivel e imprimir todos los Nodes con valores de recuento [] como impares para obtener el nivel con Nodes de números impares.
- Iterar desde el primer nivel hasta el último nivel e imprimir todos los Nodes con valores de conteo [] como incluso para obtener el nivel con Nodes de números pares.
A continuación se muestra la implementación del enfoque anterior:
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;
}
Java
// Java program to print all levels
// with odd and even number of nodes
import 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 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)
# print odd 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 print
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
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>
Producción:
The levels with odd number of nodes are: 1 3 4 The levels with even number of nodes are: 2
Tiempo Complejidad : O(N)
Espacio Auxiliar : O(N)