Encuentra el Node cuya diferencia absoluta con X da el valor máximo

Dado un árbol, y los pesos de todos los Nodes y un número entero x , la tarea es encontrar un Node i tal que |weight[i] – x| es máximo.
Ejemplos: 
 

Aporte: 
 

x = 15 
Salida: 1 
Node 1: |5 – 15| = 10 
Node 2: |10 – 15| = 5 
Node 3: |11 -15| = 4 
Node 4: |8 – 15| = 7 
Node 5: |6 -15| = 9 
 

Enfoque: Realice dfs en el árbol y realice un seguimiento del Node cuya diferencia absoluta ponderada con x da el valor máximo.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int maximum = INT_MIN, x, ans;
 
vector<int> graph[100];
vector<int> weight(100);
 
// Function to perform dfs to find
// the maximum value
void dfs(int node, int parent)
{
    // If current value is more than
    // the current maximum
    if (maximum < abs(weight[node] - x)) {
        maximum = abs(weight[node] - x);
        ans = node;
    }
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
    x = 15;
 
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
 
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
 
    dfs(1, 1);
 
    cout << ans;
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
static int maximum = Integer.MIN_VALUE, x, ans;
 
static Vector<Vector<Integer>> graph=new Vector<Vector<Integer>>();
static Vector<Integer> weight=new Vector<Integer>();
 
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
    // If current value is more than
    // the current maximum
    if (maximum < Math.abs(weight.get(node) - x))
    {
        maximum = Math.abs(weight.get(node) - x);
        ans = node;
    }
    for (int i = 0; i < graph.get(node).size(); i++)
    {
        if (graph.get(node).get(i) == parent)
            continue;
        dfs(graph.get(node).get(i), node);
    }
}
 
// Driver code
public static void main(String args[])
{
    x = 15;
 
    // Weights of the node
    weight.add(0);
    weight.add(5);
    weight.add(10);;
    weight.add(11);;
    weight.add(8);
    weight.add(6);
     
    for(int i = 0; i < 100; i++)
    graph.add(new Vector<Integer>());
 
    // Edges of the tree
    graph.get(1).add(2);
    graph.get(2).add(3);
    graph.get(2).add(4);
    graph.get(1).add(5);
 
    dfs(1, 1);
 
    System.out.println( ans);
}
}
 
// This code is contributed by Arnab Kundu

# Python implementation of the approach
from sys import maxsize
 
# Function to perform dfs to find
# the minimum value
def dfs(node, parent):
    global minimum, graph, weight, x, ans
 
    # If current value is less than
    # the current minimum
    if minimum < abs(weight[node] - x):
        minimum = abs(weight[node] - x)
        ans = node
 
    for to in graph[node]:
        if to == parent:
            continue
        dfs(to, node)
 
# Driver Code
if __name__ == "__main__":
    minimum = -maxsize
    graph = [[] for i in range(100)]
    weight = [0] * 100
    x = 15
    ans = 0
 
    # Weights of the node
    weight[1] = 5
    weight[2] = 10
    weight[3] = 11
    weight[4] = 8
    weight[5] = 6
 
    # Edges of the tree
    graph[1].append(2)
    graph[2].append(3)
    graph[2].append(4)
    graph[1].append(5)
 
    dfs(1, 1)
 
    print(ans)
 
# This code is contributed by
# sanjeev2552

// C# implementation of the approach
using System;
using System.Collections.Generic;
     
class GFG
{
 
static int maximum = int.MinValue, x, ans;
 
static List<List<int>> graph = new List<List<int>>();
static List<int> weight = new List<int>();
 
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
    // If current value is more than
    // the current maximum
    if (maximum < Math.Abs(weight[node] - x))
    {
        maximum = Math.Abs(weight[node] - x);
        ans = node;
    }
    for (int i = 0; i < graph[node].Count; i++)
    {
        if (graph[node][i] == parent)
            continue;
        dfs(graph[node][i], node);
    }
}
 
// Driver code
public static void Main(String []args)
{
    x = 15;
 
    // Weights of the node
    weight.Add(0);
    weight.Add(5);
    weight.Add(10);;
    weight.Add(11);;
    weight.Add(8);
    weight.Add(6);
     
    for(int i = 0; i < 100; i++)
    graph.Add(new List<int>());
 
    // Edges of the tree
    graph[1].Add(2);
    graph[2].Add(3);
    graph[2].Add(4);
    graph[1].Add(5);
 
    dfs(1, 1);
 
    Console.WriteLine( ans);
}
}
 
// This code is contributed by Princi Singh

<script>
 
// Javascript implementation of the approach   
 
 
    let  maximum = Number.MIN_VALUE, x, ans;
    let  graph= [];
    let  weight=[];
     
    // Function to perform dfs to find
    // the maximum value
    function dfs(node,parent)
    {
        // If current value is more than
        // the current maximum
        if (maximum < Math.abs(weight[node] - x))
        {
            maximum = Math.abs(weight[node] - x);
            ans = node;
        }
        for (let i = 0; i < graph[node].length; i++)
        {
            if (graph[node][i] == parent)
                continue;
            dfs(graph[node][i], node);
        }
    }
     
    // Driver code
    x = 15;
   
    // Weights of the node
    weight.push(0);
    weight.push(5);
    weight.push(10);;
    weight.push(11);;
    weight.push(8);
    weight.push(6);
       
    for(let i = 0; i < 100; i++)
        graph.push([]);
   
    // Edges of the tree
    graph[1].push(2);
    graph[2].push(3);
    graph[2].push(4);
    graph[1].push(5);
   
    dfs(1, 1);
   
    document.write( ans);
     
         
    // This code is contributed by unknown2108
     
</script>

1

Análisis de Complejidad: 
 

• Complejidad temporal: O(N). 
  En dfs, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida a dfs es O(N) si hay un total de N Nodes en el árbol. Por lo tanto, la complejidad del tiempo es O(N).
• Espacio Auxiliar : O(1). 
  No se requiere ningún espacio adicional, por lo que la complejidad del espacio es constante.