Dado un árbol de búsqueda binario , la tarea es encontrar el Node con el valor máximo.
Ejemplos:
Aporte:
Salida: 22
Enfoque: Simplemente atraviese el Node desde la raíz a la derecha recursivamente hasta que la derecha sea NULL. El Node cuyo derecho es NULL es el Node con el valor máximo.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node {
int data;
struct node* left;
struct node* right;
};
/* Helper function that allocates a new node
with the given data and NULL left and right
pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
struct node* insert(struct node* node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == NULL)
return (newNode(data));
else {
/* 2. Otherwise, recur down the tree */
if (data <= node->data)
node->left = insert(node->left, data);
else
node->right = insert(node->right, data);
/* return the (unchanged) node pointer */
return node;
}
}
// Function to return the minimum node
// in the given binary search tree
int maxValue(struct node* node)
{
if (node->right == NULL)
return node->data;
return maxValue(node->right);
}
// Driver code
int main()
{
// Create the BST
struct node* root = NULL;
root = insert(root, 4);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 6);
insert(root, 5);
cout << maxValue(root);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
/* A binary tree node has data,
pointer to left child and a
pointer to right child */
static class node
{
int data;
node left;
node right;
};
/* Helper function that allocates a new node
with the given data and null left and right
pointers. */
static node newNode(int data)
{
node node = new node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
static node insert(node node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null)
return (newNode(data));
else
{
/* 2. Otherwise, recur down the tree */
if (data <= node.data)
node.left = insert(node.left, data);
else
node.right = insert(node.right, data);
/* return the (unchanged) node pointer */
return node;
}
}
// Function to return the minimum node
// in the given binary search tree
static int maxValue(node node)
{
if (node.right == null)
return node.data;
return maxValue(node.right);
}
// Driver code
public static void main(String args[])
{
// Create the BST
node root = null;
root = insert(root, 4);
root = insert(root, 2);
root = insert(root, 1);
root = insert(root, 3);
root = insert(root, 6);
root = insert(root, 5);
System.out.println(maxValue(root));
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 implementation of the approach # A binary tree node has data, # pointer to left child and # a pointer to right child # Linked list node class Node: def __init__(self, data): self.data = data self.left = None self.right = None # Helper function that allocates # a new node with the given data # and None left and right pointers. def newNode(data): node = Node(0) node.data = data node.left = None node.right = None return (node) # Give a binary search tree and a number, # inserts a new node with the given number in # the correct place in the tree. Returns the new # root pointer which the caller should then use # (the standard trick to avoid using reference # parameters). def insert(node,data): # 1. If the tree is empty, # return a new, single node if (node == None): return (newNode(data)) else : # 2. Otherwise, recur down the tree if (data <= node.data): node.left = insert(node.left, data) else: node.right = insert(node.right, data) # return the (unchanged) node pointer */ return node # Function to return the minimum node # in the given binary search tree def maxValue(node): if (node.right == None): return node.data return maxValue(node.right) # Driver Code if __name__=='__main__': # Create the BST root = None root = insert(root, 4) root = insert(root, 2) root = insert(root, 1) root = insert(root, 3) root = insert(root, 6) root = insert(root, 5) print (maxValue(root)) # This code is contributed by Arnab Kundu
C#
// C# implementation of the approach
using System;
class GFG
{
/* A binary tree node has data,
pointer to left child and a
pointer to right child */
public class node
{
public int data;
public node left;
public node right;
};
/* Helper function that allocates a new node
with the given data and null left and right
pointers. */
static node newNode(int data)
{
node node = new node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
static node insert(node node, int data)
{
/* 1. If the tree is empty, return a new,
single node */
if (node == null)
return (newNode(data));
else
{
/* 2. Otherwise, recur down the tree */
if (data <= node.data)
node.left = insert(node.left, data);
else
node.right = insert(node.right, data);
/* return the (unchanged) node pointer */
return node;
}
}
// Function to return the minimum node
// in the given binary search tree
static int maxValue(node node)
{
if (node.right == null)
return node.data;
return maxValue(node.right);
}
// Driver code
public static void Main(String []args)
{
// Create the BST
node root = null;
root = insert(root, 4);
root = insert(root, 2);
root = insert(root, 1);
root = insert(root, 3);
root = insert(root, 6);
root = insert(root, 5);
Console.WriteLine(maxValue(root));
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
<script>
// Javascript implementation of the approach
/* A binary tree node has data,
pointer to left child and a
pointer to right child */
class node
{
constructor(data)
{
this.left = null;
this.right = null;
this.data = data;
}
}
/* Helper function that allocates a new node
with the given data and null left and right
pointers. */
function newNode(data)
{
let Node = new node(data);
return (Node);
}
/* Give a binary search tree and a number,
inserts a new node with the given number in
the correct place in the tree. Returns the new
root pointer which the caller should then use
(the standard trick to avoid using reference
parameters). */
function insert(Node, data)
{
/* 1. If the tree is empty, return a new,
single node */
if (Node == null)
return (newNode(data));
else
{
/* 2. Otherwise, recur down the tree */
if (data <= Node.data)
Node.left = insert(Node.left, data);
else
Node.right = insert(Node.right, data);
/* Return the (unchanged) node pointer */
return Node;
}
}
// Function to return the minimum node
// in the given binary search tree
function maxValue(Node)
{
if (Node.right == null)
return Node.data;
return maxValue(Node.right);
}
// Driver code
// Create the BST
let root = null;
root = insert(root, 4);
root = insert(root, 2);
root = insert(root, 1);
root = insert(root, 3);
root = insert(root, 6);
root = insert(root, 5);
document.write(maxValue(root));
// This code is contributed by divyeshrabadiya07
</script>
Producción:
6
Complejidad de tiempo: O (n), el peor de los casos ocurre con árboles sesgados a la derecha.
Espacio auxiliar : O(n), el número máximo de marcos de pila que podrían estar presentes en la memoria es ‘n’