K-th ancestro de un Node en Binary Tree | conjunto 3

Posted on by Rudeus Greyrat

Dado un árbol binario en el que los Nodes están numerados del 1 al N. Dado un Node y un entero positivo K . Tenemos que imprimir el ancestro Kth del Node dado en el árbol binario. Si no existe ningún ancestro de este tipo, imprima -1 .
Por ejemplo , en el siguiente árbol binario, el segundo antepasado de los Nodes 4 y 5 es 1 . El tercer ancestro del Node 4 será -1 .
 

Enfoque: Primero, encontramos la ruta de los datos clave dados desde la raíz y la almacenaremos en un vector , luego simplemente devolveremos el k -ésimo índice del vector desde el último. 
A continuación se muestra la implementación del enfoque anterior:
 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of Tree
struct node {
    node *left, *right;
    int data;
};
 
// To create a new node
node* newNode(int data)
{
    node* temp = new node;
    temp->left = temp->right = NULL;
    temp->data = data;
    return temp;
}
 
// Function to find the path from
// root to the target node
bool RootToNode(node* root, int key, vector<int>& v)
{
    if (root == NULL)
        return false;
 
    // Add current node to the path
    v.push_back(root->data);
 
    // If current node is the target node
    if (root->data == key)
        return true;
 
    // If the target node exists in
    // the left or the right sub-tree
    if (RootToNode(root->left, key, v)
        || RootToNode(root->right, key, v))
        return true;
 
    // Remove the last inserted node as
    // it is not a part of the path
    // from root to target
    v.pop_back();
    return false;
}
 
// Driver code
int main()
{
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
 
    // Given node
    int target = 4;
 
    // Vector to store the path from
    // root to the given node
    vector<int> v;
 
    // Find the path from root to the target node
    RootToNode(root, target, v);
 
    int k = 2;
 
    // Print the Kth ancestor
    if (k > v.size() - 1 || k <= 0)
        cout << -1;
    else
        cout << v[v.size() - 1 - k];
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Structure of Tree
static class node
{
    node left, right;
    int data;
};
 
// To create a new node
static node newNode(int data)
{
    node temp = new node();
    temp.left = temp.right = null;
    temp.data = data;
    return temp;
}
 
// Function to find the path from
// root to the target node
static boolean RootToNode(node root, int key,
                            Vector<Integer> v)
{
    if (root == null)
        return false;
 
    // Add current node to the path
    v.add(root.data);
 
    // If current node is the target node
    if (root.data == key)
        return true;
 
    // If the target node exists in
    // the left or the right sub-tree
    if (RootToNode(root.left, key, v)
        || RootToNode(root.right, key, v))
        return true;
 
    // Remove the last inserted node as
    // it is not a part of the path
    // from root to target
    v.removeElementAt(v.size()-1);
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
 
    // Given node
    int target = 4;
 
    // Vector to store the path from
    // root to the given node
    Vector<Integer> v = new Vector<>();
 
    // Find the path from root to the target node
    RootToNode(root, target, v);
 
    int k = 2;
 
    // Print the Kth ancestor
    if (k > v.size() - 1 || k <= 0)
        System.out.println(-1);
    else
        System.out.println(v.get(v.size() - 1 - k));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation of the approach
 
# To create a  node
class Node:
  
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to find the path
# from root to the target node
def RootToNode(root, key, v):
  
    if root == None:
        return False
 
    # Add current node to the path
    v.append(root.data)
 
    # If current node is the target node
    if root.data == key:
        return True
 
    # If the target node exists in
    # the left or the right sub-tree
    if (RootToNode(root.left, key, v) or
       RootToNode(root.right, key, v)):
        return True
 
    # Remove the last inserted node
    # as it is not a part of the
    # path from root to target
    v.pop()
    return False
  
# Driver code
if __name__ == "__main__":
  
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    root.right.left = Node(6)
    root.right.right = Node(7)
 
    # Given node
    target, k = 4, 2
 
    # Vector to store the path
    # from root to the given node
    v = []
 
    # Find the path from root to the target node
    RootToNode(root, target, v)
 
    # Print the Kth ancestor
    if k > len(v) - 1 or k <= 0:
        print(-1)
    else:
        print(v[len(v) - 1 - k])
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of above approach
using System.Collections.Generic;
using System;
 
class GFG
{
 
// Structure of Tree
public class node
{
    public node left, right;
    public int data;
};
 
// To create a new node
static node newNode(int data)
{
    node temp = new node();
    temp.left = temp.right = null;
    temp.data = data;
    return temp;
}
 
// Function to find the path from
// root to the target node
static bool RootToNode(node root, int key,
                            List<int> v)
{
    if (root == null)
        return false;
 
    // Add current node to the path
    v.Add(root.data);
 
    // If current node is the target node
    if (root.data == key)
        return true;
 
    // If the target node exists in
    // the left or the right sub-tree
    if (RootToNode(root.left, key, v)
        || RootToNode(root.right, key, v))
        return true;
 
    // Remove the last inserted node as
    // it is not a part of the path
    // from root to target
    v.Remove(v.Count-1);
    return false;
}
 
// Driver code
public static void Main(String[] args)
{
    node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
 
    // Given node
    int target = 4;
 
    // Vector to store the path from
    // root to the given node
    List<int> v = new List<int>();
 
    // Find the path from root to the target node
    RootToNode(root, target, v);
 
    int k = 2;
 
    // Print the Kth ancestor
    if (k > v.Count - 1 || k <= 0)
        Console.WriteLine(-1);
    else
        Console.WriteLine(v[v.Count - 1 - k]);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// JavaScript implementation of above approach
 
// Structure of Tree
class node
{
    constructor()
    {
        this.left = null;
        this.right = null;
        this.data = 0;
    }
};
 
// To create a new node
function newNode(data)
{
    var temp = new node();
    temp.left = temp.right = null;
    temp.data = data;
    return temp;
}
 
// Function to find the path from
// root to the target node
function RootToNode(root, key, v)
{
    if (root == null)
        return false;
 
    // push current node to the path
    v.push(root.data);
 
    // If current node is the target node
    if (root.data == key)
        return true;
 
    // If the target node exists in
    // the left or the right sub-tree
    if (RootToNode(root.left, key, v)
        || RootToNode(root.right, key, v))
        return true;
 
    // Remove the last inserted node as
    // it is not a part of the path
    // from root to target
    v.pop();
    return false;
}
 
// Driver code
var root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(6);
root.right.right = newNode(7);
// Given node
var target = 4;
// Vector to store the path from
// root to the given node
var v = [];
// Find the path from root to the target node
RootToNode(root, target, v);
var k = 2;
// Print the Kth ancestor
if (k > v.length - 1 || k <= 0)
    document.write(-1);
else
    document.write(v[v.length - 1 - k]);
 
</script>
Producción: 

1

 

Publicación traducida automáticamente

Artículo escrito por iamAAMIR y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

Deja una respuesta

Tu dirección de correo electrónico no será publicada. Los campos obligatorios están marcados con *