Dado un árbol binario que consta de N Nodes, la tarea es dividir el árbol binario en dos subárboles eliminando un borde de modo que se minimice la diferencia absoluta de la suma de los subárboles.
Ejemplo:
Entrada: 1
/ \
2 3
/ \ \
4 5 8
/ \
6 7
Salida: 6
Explicación: Divide el árbol entre los vértices 3 y 8. Los subárboles creados tienen sumas 21 y 15Entrada : 1
/ \
2 3
/ \ / \
4 5 6 7
Salida : 4
Explicación: Divide el árbol entre los vértices 1 y 3. Los subárboles creados tienen sumas 16 y 12
Enfoque: el problema dado se puede resolver siguiendo los pasos a continuación:
- Inicialice la variable minDiff al valor máximo de Integer que almacenará la respuesta
- Use el recorrido posterior al pedido para almacenar la suma del Node actual, el subárbol izquierdo y el subárbol derecho en el Node actual
- Use el recorrido de preorden y en cada llamada recursiva encuentre la suma de los subárboles si la división es entre:
- Node actual y Node secundario izquierdo
- Node actual y Node secundario derecho
- Actualice minDiff si en cualquier llamada recursiva la diferencia de subárboles es menor que minDiff
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
struct Node {
Node* left;
Node* right;
int val;
Node(int val)
{
this->val = val;
this->left = nullptr;
this->right = nullptr;
}
};
int postOrder(Node* root)
{
if (root == nullptr)
return 0;
root->val += postOrder(root->left) + postOrder(root->right);
return root->val;
}
void preOrder(Node* root, int minDiff[])
{
if (root == nullptr)
return;
// Absolute difference in subtrees
// if left edge is broken
int leftDiff = abs(root->left->val - (root->val - root->left->val));
// Absolute difference in subtrees
// if right edge is broken
int rightDiff = abs(root->right->val - (root->val - root->right->val));
// Update minDiff if a difference
// lesser than it is found
minDiff[0] = min(minDiff[0], min(leftDiff, rightDiff));
return;
}
// Function to calculate minimum absolute
// difference of subtrees after
// splitting the tree into two parts
int minAbsDiff(Node* root)
{
// Reference variable
// to store the answer
int minDiff[1];
minDiff[0] = INT_MAX;
// Function to store sum of values
// of current node, left subtree and
// right subtree in place of
// current node's value
postOrder(root);
// Function to perform every
// possible split and calculate
// absolute difference of subtrees
preOrder(root, minDiff);
return minDiff[0];
}
int main()
{
// Construct the tree
Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);
root->right->left = new Node(6);
root->right->right = new Node(7);
// Print the output
cout << minAbsDiff(root);
return 0;
}
// This code is contributed by mukesh07.
Java
// Java implementation for the above approach
import java.lang.Math;
import java.io.*;
class GFG {
static class Node {
Node left, right;
int val;
public Node(int val)
{
this.val = val;
this.left = this.right = null;
}
}
// Function to calculate minimum absolute
// difference of subtrees after
// splitting the tree into two parts
public static int minAbsDiff(Node root)
{
// Reference variable
// to store the answer
int[] minDiff = new int[1];
minDiff[0] = Integer.MAX_VALUE;
// Function to store sum of values
// of current node, left subtree and
// right subtree in place of
// current node's value
postOrder(root);
// Function to perform every
// possible split and calculate
// absolute difference of subtrees
preOrder(root, minDiff);
return minDiff[0];
}
public static int postOrder(Node root)
{
if (root == null)
return 0;
root.val += postOrder(root.left)
+ postOrder(root.right);
return root.val;
}
public static void preOrder(Node root,
int[] minDiff)
{
if (root == null)
return;
// Absolute difference in subtrees
// if left edge is broken
int leftDiff
= Math.abs(root.left.val
- (root.val - root.left.val));
// Absolute difference in subtrees
// if right edge is broken
int rightDiff
= Math.abs(root.right.val
- (root.val - root.right.val));
// Update minDiff if a difference
// lesser than it is found
minDiff[0]
= Math.min(minDiff[0],
Math.min(leftDiff, rightDiff));
return;
}
// Driver code
public static void main(String[] args)
{
// Construct the tree
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
// Print the output
System.out.println(minAbsDiff(root));
}
}
Python3
# Python3 implementation for the above approach import sys # Structure of Node class Node: def __init__(self, val): self.val = val self.left = None self.right = None # Function to calculate minimum absolute # difference of subtrees after # splitting the tree into two parts def minAbsDiff(root): # Reference variable # to store the answer minDiff = [0]*(1) minDiff[0] = sys.maxsize # Function to store sum of values # of current node, left subtree and # right subtree in place of # current node's value postOrder(root) # Function to perform every # possible split and calculate # absolute difference of subtrees preOrder(root, minDiff) return minDiff[0] def postOrder(root): if (root == None): return 0 root.val += postOrder(root.left) + postOrder(root.right) return root.val def preOrder(root, minDiff): if (root == None): return # Absolute difference in subtrees # if left edge is broken leftDiff = abs(root.left.val - (root.val - root.left.val)) # Absolute difference in subtrees # if right edge is broken rightDiff = abs(root.right.val - (root.val - root.right.val)) # Update minDiff if a difference # lesser than it is found minDiff[0] = min(minDiff[0], min(leftDiff, rightDiff)) return # Construct the tree 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) # Print the output print(minAbsDiff(root)) # This code is contributed by rameshtravel07.
C#
// C# implementation for the above approach
using System;
public class GFG {
class Node {
public Node left, right;
public int val;
}
static Node newNode(int data)
{
Node newNode = new Node();
newNode.val = data;
newNode.left= null;
newNode.right = null;
return newNode;
}
// Function to calculate minimum absolute
// difference of subtrees after
// splitting the tree into two parts
static int minAbsDiff(Node root)
{
// Reference variable
// to store the answer
int[] minDiff = new int[1];
minDiff[0] = Int32.MaxValue;
// Function to store sum of values
// of current node, left subtree and
// right subtree in place of
// current node's value
postOrder(root);
// Function to perform every
// possible split and calculate
// absolute difference of subtrees
preOrder(root, minDiff);
return minDiff[0];
}
static int postOrder(Node root)
{
if (root == null)
return 0;
root.val += postOrder(root.left)
+ postOrder(root.right);
return root.val;
}
static void preOrder(Node root,
int[] minDiff)
{
if (root == null)
return;
// Absolute difference in subtrees
// if left edge is broken
int leftDiff
= Math.Abs(root.left.val
- (root.val - root.left.val));
// Absolute difference in subtrees
// if right edge is broken
int rightDiff
= Math.Abs(root.right.val
- (root.val - root.right.val));
// Update minDiff if a difference
// lesser than it is found
minDiff[0]
= Math.Min(minDiff[0],
Math.Min(leftDiff, rightDiff));
return;
}
// Driver code
public static void Main()
{
// Construct the tree
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);
// Print the output
Console.Write(minAbsDiff(root));
}
}
// This code is contributed by SURENDRA_GANGWAR.
Javascript
<script>
// Javascript implementation for the above approach
// Structure of Node
class Node
{
constructor(val) {
this.left = null;
this.right = null;
this.val = val;
}
}
// Function to calculate minimum absolute
// difference of subtrees after
// splitting the tree into two parts
function minAbsDiff(root)
{
// Reference variable
// to store the answer
let minDiff = new Array(1);
minDiff[0] = Number.MAX_VALUE;
// Function to store sum of values
// of current node, left subtree and
// right subtree in place of
// current node's value
postOrder(root);
// Function to perform every
// possible split and calculate
// absolute difference of subtrees
preOrder(root, minDiff);
return minDiff[0];
}
function postOrder(root)
{
if (root == null)
return 0;
root.val += postOrder(root.left)
+ postOrder(root.right);
return root.val;
}
function preOrder(root, minDiff)
{
if (root == null)
return;
// Absolute difference in subtrees
// if left edge is broken
let leftDiff
= Math.abs(root.left.val
- (root.val - root.left.val));
// Absolute difference in subtrees
// if right edge is broken
let rightDiff
= Math.abs(root.right.val
- (root.val - root.right.val));
// Update minDiff if a difference
// lesser than it is found
minDiff[0]
= Math.min(minDiff[0],
Math.min(leftDiff, rightDiff));
return;
}
// Construct the tree
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
// Print the output
document.write(minAbsDiff(root));
// This code is contributed by decode2207.
</script>
Producción:
4
Complejidad de Tiempo : O(N)
Espacio Auxiliar : O(1)
Publicación traducida automáticamente
Artículo escrito por koladiyadrashti123 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA