Dado un árbol binario que consiste solo en 0 y 1 , la tarea es imprimir el recuento de niveles en el árbol binario en el que todos los 1 se colocan consecutivamente en un solo grupo.
Ejemplos:
Entrada: 0
/ \
1 0
/ \ / \
1 0 1 0
Salida: 2
Explicación: En los Niveles 1 y 2, todos los Nodes con valor 1 se colocan consecutivamente.
Entrada: 0
/ \
1 0
/ \ \
1 1 0
/ \ \ / \
1 1 1 0 0
Salida: 4
Explicación: En todos los niveles, los Nodes con valor 1 se colocan consecutivamente.
Enfoque: siga los pasos a continuación para resolver el problema:
- Realice el cruce de orden de nivel utilizando Queue .
- Recorra cada nivel del árbol binario y considere las siguientes tres variables:
- flag1: se establece en 1 después de la primera aparición del Node con valor 1 .
- flag0: se establece en 1 después de la primera aparición del Node con valor 0 después de la aparición de cualquier Node con valor 1.
- flag2: se establece después de la primera aparición del Node con el valor 1 después de que flag0 y flag1 se establezcan en 1 .
- Después de atravesar cada nivel, verifique si flag1 está establecido en 1 y flag2 es 0. Si es cierto, incluya ese nivel en el conteo.
- Finalmente, imprima el conteo obtenido.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// A Binary Tree Node
struct node {
// Left Child
struct node* left;
int data;
// Right Child
struct node* right;
};
// Function to perform level order traversal
// count levels with all 1s grouped together
int countLevels(node* root)
{
if (root == NULL)
return 0;
int Count = 0;
// Create an empty queue for
// level order traversal
queue<node*> q;
// Stores front element of the queue
node* curr;
// Enqueue root and NULL node
q.push(root);
// Stores Nodes of
// Current Level
while (!q.empty()) {
int n = q.size();
int flag0 = 0, flag1 = 0, flag2 = 0;
while (n--) {
// Stores first node of
// the current level
curr = q.front();
q.pop();
if (curr) {
// If left child exists
if (curr->left)
// Push into the Queue
q.push(curr->left);
// If right child exists
if (curr->right)
// Push into the Queue
q.push(curr->right);
if (curr->data == 1) {
// If current node is the first
// node with value 1
if (!flag1)
flag1 = 1;
// If current node has value 1
// after occurrence of nodes
// with value 0 following a
// sequence of nodes with value 1
if (flag1 && flag0)
flag2 = 1;
}
// If current node is the first node
// with value 0 after a sequence
// of nodes with value 1
if (curr->data == 0 && flag1)
flag0 = 1;
}
}
if (flag1 && !flag2)
Count++;
}
return Count;
}
// Function to create a Tree Node
node* newNode(int data)
{
node* temp = new node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Driver Code
int main()
{
node* root = newNode(0);
root->left = newNode(0);
root->right = newNode(1);
root->left->left = newNode(0);
root->left->right = newNode(1);
root->right->left = newNode(1);
root->right->right = newNode(0);
cout << countLevels(root);
return 0;
}
Java
// Java Program to implement
// the above approach
import java.util.*;
class GFG
{
// A Binary Tree Node
static class node
{
// Left Child
node left;
int data;
// Right Child
node right;
};
// Function to perform level order traversal
// count levels with all 1s grouped together
static int countLevels(node root)
{
if (root == null)
return 0;
int Count = 0;
// Create an empty queue for
// level order traversal
Queue<node> q = new LinkedList<>();
// Stores front element of the queue
node curr;
// Enqueue root and null node
q.add(root);
// Stores Nodes of
// Current Level
while (!q.isEmpty())
{
int n = q.size();
int flag0 = 0, flag1 = 0, flag2 = 0;
while (n-- >0)
{
// Stores first node of
// the current level
curr = q.peek();
q.remove();
if (curr != null)
{
// If left child exists
if (curr.left != null)
// Push into the Queue
q.add(curr.left);
// If right child exists
if (curr.right != null)
// Push into the Queue
q.add(curr.right);
if (curr.data == 1)
{
// If current node is the first
// node with value 1
if (flag1 == 0)
flag1 = 1;
// If current node has value 1
// after occurrence of nodes
// with value 0 following a
// sequence of nodes with value 1
if (flag1 > 0 && flag0 > 0)
flag2 = 1;
}
// If current node is the first node
// with value 0 after a sequence
// of nodes with value 1
if (curr.data == 0 && flag1 > 0)
flag0 = 1;
}
}
if (flag1 > 0 && flag2 == 0)
Count++;
}
return Count;
}
// Function to create a Tree Node
static node newNode(int data)
{
node temp = new node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
// Driver Code
public static void main(String[] args)
{
node root = newNode(0);
root.left = newNode(0);
root.right = newNode(1);
root.left.left = newNode(0);
root.left.right = newNode(1);
root.right.left = newNode(1);
root.right.right = newNode(0);
System.out.print(countLevels(root));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to implement # the above approach from collections import deque # A Binary Tree Node class node: def __init__(self): # Left Child self.left = None self.data = 0 # Right Child self.right = None # Function to perform level order traversal # count levels with all 1s grouped together def countLevels(root): if (root == None): return 0 Count = 0 # Create an empty queue for # level order traversal q = deque() # Stores front element of the queue curr = node() # Enqueue root and None node q.append(root) # Stores Nodes of # Current Level while q: n = len(q) flag0 = 0 flag1 = 0 flag2 = 0 while (n): # Stores first node of # the current level curr = q[0] q.popleft() if (curr): # If left child exists if (curr.left): # Push into the Queue q.append(curr.left) # If right child exists if (curr.right): # Push into the Queue q.append(curr.right) if (curr.data == 1): # If current node is the first # node with value 1 if (not flag1): flag1 = 1 # If current node has value 1 # after occurrence of nodes # with value 0 following a # sequence of nodes with value 1 if (flag1 and flag0): flag2 = 1 # If current node is the first node # with value 0 after a sequence # of nodes with value 1 if (curr.data == 0 and flag1): flag0 = 1 n -= 1 if (flag1 and not flag2): Count += 1 return Count # Function to create a Tree Node def newNode(data): temp = node() temp.data = data temp.left = None temp.right = None return temp # Driver Code if __name__ == "__main__": root = newNode(0) root.left = newNode(0) root.right = newNode(1) root.left.left = newNode(0) root.left.right = newNode(1) root.right.left = newNode(1) root.right.right = newNode(0) print(countLevels(root)) # This code is contributed by sanjeev2552
C#
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// A Binary Tree Node
class node
{
// Left Child
public node left;
public int data;
// Right Child
public node right;
};
// Function to perform level order traversal
// count levels with all 1s grouped together
static int countLevels(node root)
{
if (root == null)
return 0;
int Count = 0;
// Create an empty queue for
// level order traversal
Queue<node> q = new Queue<node>();
// Stores front element of the queue
node curr;
// Enqueue root and null node
q.Enqueue(root);
// Stores Nodes of
// Current Level
while (q.Count != 0)
{
int n = q.Count;
int flag0 = 0, flag1 = 0, flag2 = 0;
while (n-- >0)
{
// Stores first node of
// the current level
curr = q.Peek();
q.Dequeue();
if (curr != null)
{
// If left child exists
if (curr.left != null)
// Push into the Queue
q.Enqueue(curr.left);
// If right child exists
if (curr.right != null)
// Push into the Queue
q.Enqueue(curr.right);
if (curr.data == 1)
{
// If current node is the first
// node with value 1
if (flag1 == 0)
flag1 = 1;
// If current node has value 1
// after occurrence of nodes
// with value 0 following a
// sequence of nodes with value 1
if (flag1 > 0 && flag0 > 0)
flag2 = 1;
}
// If current node is the first node
// with value 0 after a sequence
// of nodes with value 1
if (curr.data == 0 && flag1 > 0)
flag0 = 1;
}
}
if (flag1 > 0 && flag2 == 0)
Count++;
}
return Count;
}
// Function to create a Tree Node
static node newNode(int data)
{
node temp = new node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
// Driver Code
public static void Main(String[] args)
{
node root = newNode(0);
root.left = newNode(0);
root.right = newNode(1);
root.left.left = newNode(0);
root.left.right = newNode(1);
root.right.left = newNode(1);
root.right.right = newNode(0);
Console.Write(countLevels(root));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript program for above approach
// A Binary Tree Node
class node
{
constructor(data) {
this.left = null;
this.right = null;
this.data = data;
}
}
// Function to create a Tree Node
function newNode(data)
{
let temp = new node(data);
return temp;
}
// Function to perform level order traversal
// count levels with all 1s grouped together
function countLevels(root)
{
if (root == null)
return 0;
let Count = 0;
// Create an empty queue for
// level order traversal
let q = [];
// Stores front element of the queue
let curr;
// Enqueue root and null node
q.push(root);
// Stores Nodes of
// Current Level
while (q.length > 0)
{
let n = q.length;
let flag0 = 0, flag1 = 0, flag2 = 0;
while (n-- >0)
{
// Stores first node of
// the current level
curr = q[0];
q.shift();
if (curr != null)
{
// If left child exists
if (curr.left != null)
// Push into the Queue
q.push(curr.left);
// If right child exists
if (curr.right != null)
// Push into the Queue
q.push(curr.right);
if (curr.data == 1)
{
// If current node is the first
// node with value 1
if (flag1 == 0)
flag1 = 1;
// If current node has value 1
// after occurrence of nodes
// with value 0 following a
// sequence of nodes with value 1
if (flag1 > 0 && flag0 > 0)
flag2 = 1;
}
// If current node is the first node
// with value 0 after a sequence
// of nodes with value 1
if (curr.data == 0 && flag1 > 0)
flag0 = 1;
}
}
if (flag1 > 0 && flag2 == 0)
Count++;
}
return Count;
}
let root = newNode(0);
root.left = newNode(0);
root.right = newNode(1);
root.left.left = newNode(0);
root.left.right = newNode(1);
root.right.left = newNode(1);
root.right.right = newNode(0);
document.write(countLevels(root));
</script>
Producción:
2
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por rohitsingh07052 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA