Dado un árbol binario, encuentre la suma máxima de caminos. La ruta puede comenzar y terminar en cualquier Node del árbol.
Ejemplo:
Input: Root of below tree
1
/ \
2 3
Output: 6
See below diagram for another example.
1+2+3
Para cada Node, puede haber cuatro formas en que la ruta máxima atraviesa el Node:
1. Solo Node
2. Ruta máxima a través del elemento secundario izquierdo + Node
3. Ruta máxima a través del elemento secundario derecho + Node
4. Ruta máxima a través del elemento secundario izquierdo + Node + Máx. camino a través del niño derecho
La idea es seguir el rastro de cuatro caminos y recoger el máximo al final. Una cosa importante a tener en cuenta es que la raíz de cada subárbol debe devolver la suma máxima de la ruta de modo que, como máximo, esté involucrado un hijo de la raíz. Esto es necesario para la llamada a la función principal. En el siguiente código, esta suma se almacena en ‘max_single’ y la función recursiva la devuelve.
C++
// C/C++ program to find maximum path sum in Binary Tree
#include<bits/stdc++.h>
using namespace std;
// A binary tree node
struct Node
{
int data;
struct Node* left, *right;
};
// A utility function to allocate a new node
struct Node* newNode(int data)
{
struct Node* newNode = new Node;
newNode->data = data;
newNode->left = newNode->right = NULL;
return (newNode);
}
// This function returns overall maximum path sum in 'res'
// And returns max path sum going through root.
int findMaxUtil(Node* root, int &res)
{
//Base Case
if (root == NULL)
return 0;
// l and r store maximum path sum going through left and
// right child of root respectively
int l = findMaxUtil(root->left,res);
int r = findMaxUtil(root->right,res);
// Max path for parent call of root. This path must
// include at-most one child of root
int max_single = max(max(l, r) + root->data, root->data);
// Max Top represents the sum when the Node under
// consideration is the root of the maxsum path and no
// ancestors of root are there in max sum path
int max_top = max(max_single, l + r + root->data);
res = max(res, max_top); // Store the Maximum Result.
return max_single;
}
// Returns maximum path sum in tree with given root
int findMaxSum(Node *root)
{
// Initialize result
int res = INT_MIN;
// Compute and return result
findMaxUtil(root, res);
return res;
}
// Driver program
int main(void)
{
struct Node *root = newNode(10);
root->left = newNode(2);
root->right = newNode(10);
root->left->left = newNode(20);
root->left->right = newNode(1);
root->right->right = newNode(-25);
root->right->right->left = newNode(3);
root->right->right->right = newNode(4);
cout << "Max path sum is " << findMaxSum(root);
return 0;
}
Java
// Java program to find maximum path sum in Binary Tree
/* Class containing left and right child of current
node and key value*/
class Node {
int data;
Node left, right;
public Node(int item) {
data = item;
left = right = null;
}
}
// An object of Res is passed around so that the
// same value can be used by multiple recursive calls.
class Res {
public int val;
}
class BinaryTree {
// Root of the Binary Tree
Node root;
// This function returns overall maximum path sum in 'res'
// And returns max path sum going through root.
int findMaxUtil(Node node, Res res)
{
// Base Case
if (node == null)
return 0;
// l and r store maximum path sum going through left and
// right child of root respectively
int l = findMaxUtil(node.left, res);
int r = findMaxUtil(node.right, res);
// Max path for parent call of root. This path must
// include at-most one child of root
int max_single = Math.max(Math.max(l, r) + node.data,
node.data);
// Max Top represents the sum when the Node under
// consideration is the root of the maxsum path and no
// ancestors of root are there in max sum path
int max_top = Math.max(max_single, l + r + node.data);
// Store the Maximum Result.
res.val = Math.max(res.val, max_top);
return max_single;
}
int findMaxSum() {
return findMaxSum(root);
}
// Returns maximum path sum in tree with given root
int findMaxSum(Node node) {
// Initialize result
// int res2 = Integer.MIN_VALUE;
Res res = new Res();
res.val = Integer.MIN_VALUE;
// Compute and return result
findMaxUtil(node, res);
return res.val;
}
/* Driver program to test above functions */
public static void main(String args[]) {
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(2);
tree.root.right = new Node(10);
tree.root.left.left = new Node(20);
tree.root.left.right = new Node(1);
tree.root.right.right = new Node(-25);
tree.root.right.right.left = new Node(3);
tree.root.right.right.right = new Node(4);
System.out.println("maximum path sum is : " +
tree.findMaxSum());
}
}
Python
# Python program to find maximum path sum in Binary Tree
# A Binary Tree Node
class Node:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# This function returns overall maximum path sum in 'res'
# And returns max path sum going through root
def findMaxUtil(root):
# Base Case
if root is None:
return 0
# l and r store maximum path sum going through left
# and right child of root respectively
l = findMaxUtil(root.left)
r = findMaxUtil(root.right)
# Max path for parent call of root. This path
# must include at most one child of root
max_single = max(max(l, r) + root.data, root.data)
# Max top represents the sum when the node under
# consideration is the root of the maxSum path and
# no ancestor of root are there in max sum path
max_top = max(max_single, l+r+ root.data)
# Static variable to store the changes
# Store the maximum result
findMaxUtil.res = max(findMaxUtil.res, max_top)
return max_single
# Return maximum path sum in tree with given root
def findMaxSum(root):
# Initialize result
findMaxUtil.res = float("-inf")
# Compute and return result
findMaxUtil(root)
return findMaxUtil.res
# Driver program
root = Node(10)
root.left = Node(2)
root.right = Node(10);
root.left.left = Node(20);
root.left.right = Node(1);
root.right.right = Node(-25);
root.right.right.left = Node(3);
root.right.right.right = Node(4);
print "Max path sum is " ,findMaxSum(root);
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
C#
// C# program to find maximum
// path sum in Binary Tree
using System;
/* Class containing left and
right child of current
node and key value*/
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// An object of Res is passed
// around so that the same value
// can be used by multiple recursive calls.
class Res
{
public int val;
}
public class BinaryTree
{
// Root of the Binary Tree
Node root;
// This function returns overall
// maximum path sum in 'res' And
// returns max path sum going through root.
int findMaxUtil(Node node, Res res)
{
// Base Case
if (node == null)
return 0;
// l and r store maximum path
// sum going through left and
// right child of root respectively
int l = findMaxUtil(node.left, res);
int r = findMaxUtil(node.right, res);
// Max path for parent call of root.
// This path must include
// at-most one child of root
int max_single = Math.Max(Math.Max(l, r) +
node.data, node.data);
// Max Top represents the sum
// when the Node under
// consideration is the root
// of the maxsum path and no
// ancestors of root are there
// in max sum path
int max_top = Math.Max(max_single,
l + r + node.data);
// Store the Maximum Result.
res.val = Math.Max(res.val, max_top);
return max_single;
}
int findMaxSum()
{
return findMaxSum(root);
}
// Returns maximum path
// sum in tree with given root
int findMaxSum(Node node)
{
// Initialize result
// int res2 = int.MinValue;
Res res = new Res();
res.val = int.MinValue;
// Compute and return result
findMaxUtil(node, res);
return res.val;
}
/* Driver code */
public static void Main(String []args)
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(10);
tree.root.left = new Node(2);
tree.root.right = new Node(10);
tree.root.left.left = new Node(20);
tree.root.left.right = new Node(1);
tree.root.right.right = new Node(-25);
tree.root.right.right.left = new Node(3);
tree.root.right.right.right = new Node(4);
Console.WriteLine("maximum path sum is : " +
tree.findMaxSum());
}
}
// This code is contributed Rajput-Ji.
Javascript
<script>
// JavaScript program to find maximum
// path sum in Binary Tree
class Node
{
constructor(item) {
this.left = null;
this.right = null;
this.data = item;
}
}
let val;
// Root of the Binary Tree
let root;
// This function returns overall maximum path sum in 'res'
// And returns max path sum going through root.
function findMaxUtil(node)
{
// Base Case
if (node == null)
return 0;
// l and r store maximum path sum going through left and
// right child of root respectively
let l = findMaxUtil(node.left);
let r = findMaxUtil(node.right);
// Max path for parent call of root. This path must
// include at-most one child of root
let max_single = Math.max(Math.max(l, r) + node.data,
node.data);
// Max Top represents the sum when the Node under
// consideration is the root of the maxsum path and no
// ancestors of root are there in max sum path
let max_top = Math.max(max_single, l + r + node.data);
// Store the Maximum Result.
val = Math.max(val, max_top);
return max_single;
}
function findMaxsum() {
return findMaxSum(root);
}
// Returns maximum path sum in tree with given root
function findMaxSum(node) {
// Initialize result
// int res2 = Integer.MIN_VALUE;
val = Number.MIN_VALUE;
// Compute and return result
findMaxUtil(node);
return val;
}
root = new Node(10);
root.left = new Node(2);
root.right = new Node(10);
root.left.left = new Node(20);
root.left.right = new Node(1);
root.right.right = new Node(-25);
root.right.right.left = new Node(3);
root.right.right.right = new Node(4);
document.write("Max path sum is : " + findMaxsum());
</script>
Producción:
Max path sum is 42
Complejidad de tiempo: O(n) donde n es el número de Nodes en el árbol binario.
Este artículo es una contribución de Anmol Varshney (Perfil de FB: https://www.facebook.com/anmolvarshney695 ). Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA