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

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 mínimo.
Ejemplos: 
 

Aporte: 
 

x = 15 
Salida: 3 
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 mínimo.
A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
int minimum = INT_MAX, x, ans;
 
vector<int> graph[100];
vector<int> weight(100);
 
// Function to perform dfs to find
// the minimum value
void dfs(int node, int parent)
{
    // If current value is less than
    // the current minimum
    if (minimum > abs(weight[node] - x)) {
        minimum = 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.*;
import java.lang.*;
 
class GFG
{
 
    static int minimum = Integer.MAX_VALUE, x, ans;
 
    @SuppressWarnings("unchecked")
    static Vector<Integer>[] graph = new Vector[100];
    static int[] weight = new int[100];
 
    // This block is executed even before main() function
    // This is necessary otherwise this program will
    // throw "NullPointerException"
    static
    {
        for (int i = 0; i < 100; i++)
            graph[i] = new Vector<>();
    }
 
    // Function to perform dfs to find
    // the minimum xored value
    static void dfs(int node, int parent)
    {
 
        // If current value is less than
        // the current minimum
        if (minimum > Math.abs(weight[node] - x))
        {
            minimum = Math.abs(weight[node] - x);
            ans = node;
        }
        for (int to : graph[node])
        {
            if (to == parent)
                continue;
            dfs(to, node);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        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].add(2);
        graph[2].add(3);
        graph[2].add(4);
        graph[1].add(5);
 
        dfs(1, 1);
 
        System.out.println(ans);
    }
}
 
// This code is contributed by SHUBHAMSINGH10

# Python3 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 minimum = int.MaxValue, 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 minimum value
static void dfs(int node, int parent)
{
    // If current value is more than
    // the current minimum
    if (minimum > Math.Abs(weight[node] - x))
    {
        minimum = 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 shubhamsingh10

<script>
 
// Javascript implementation of the approach
     
    let minimum = Number.MAX_VALUE, x, ans;
     
    let graph = new Array(100);
     
     
    let weight = new Array(100);
    for(let i=0;i<100;i++)
    {
        graph[i]=[];
        weight[i]=0;
    }
     
    // Function to perform dfs to find
    // the minimum xored value
    function dfs(node,parent)
    {
        // If current value is less than
        // the current minimum
        if (minimum > Math.abs(weight[node] - x))
        {
            minimum = Math.abs(weight[node] - x);
            ans = node;
        }
        for(let to=0;to<graph[node].length;to++)
        {
            if(graph[node][to] == parent)
                continue
            dfs(graph[node][to], node);  
        }
         
    }
     
    // Driver code
     
    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(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>

3

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.