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 DFS en el árbol de modo que altura[Node] = 1 + altura[padre]
- Una vez que se completa el recorrido de DFS, 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 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;
}
Java
// Java program to print all levels
// with odd and even number of nodes
import java.util.*;
@SuppressWarnings("unchecked")
class GFG{
// Function for DFS in a tree
static void dfs(int node, int parent,
int []height, int []vis,
ArrayList []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(int it : (ArrayList<Integer>)tree[node])
{
// If the node is not visited
if (vis[it] == 0)
{
// Call the dfs function
dfs(it, node, height, vis, tree);
}
}
}
// 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];
Arrays.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
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.print("\nThe levels with even " +
"number of nodes are: ");
for(int i = 1; i <= maxLevel; i++)
{
if (mark[i] % 2 == 0)
{
System.out.print(i + " ");
}
}
}
// Driver code
public static void main(String []s)
{
// 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];
Arrays.fill(vis, 0);
height[0] = 0;
// Call the dfs function
dfs(1, 0, height, vis, tree);
// Function to print
printLevelsOddEven(N, vis, height);
}
}
// This code is contributed by pratham76
Python3
# Python3 program to print all levels
# with odd and even number of nodes
# Function for DFS in a tree
def dfs(node, parent, height, vis, 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 it in tree[node]:
# if the node is not visited
if not vis[it]:
# call the dfs function
dfs(it, node, height, vis, tree)
# 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 + 1):
if mark[i] % 2 == 0:
print(i, end = " ")
# Driver Code
if __name__ == "__main__":
# Construct the tree
N = 9
tree = [[] for i in range(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 dfs function
dfs(1, 0, height, vis, tree)
# Function to print
printLevelsOddEven(N, vis, height)
# This code is contributed by Rituraj Jain
C#
// C# program to print all levels
// with odd and even number of nodes
using System;
using System.Collections;
class GFG{
// Function for DFS in a tree
static void dfs(int node, int parent,
int []height, int []vis,
ArrayList []tree)
{
// Calculate the level of every node
height[node] = 1 + height[parent];
// Mark every node as visited
vis[node] = 1;
// Iterate in the subtree
foreach (int it in tree[node])
{
// If the node is not visited
if (vis[it] == 0)
{
// Call the dfs function
dfs(it, node, height, vis, tree);
}
}
}
// 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 dfs function
dfs(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 DFS in a tree
function dfs(node, parent, height, vis, 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(let it = 0; it < tree[node].length; it++)
{
// If the node is not visited
if (vis[tree[node][it]] == 0)
{
// Call the dfs function
dfs(tree[node][it], node, height, vis, tree);
}
}
}
// 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 dfs function
dfs(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)