Given a Binary Search Tree, the task is to find the node with maximum value.
Examples:
Input:
Output: 22
Approach: Just traverse the node from root to right recursively until right is NULL. The node whose right is NULL is the node with the maximum value.
Below is the implementation of the above approach:
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> |
Output:
6
Time Complexity: O(n), worst case happens for right skewed trees.
Auxiliary Space: O(n), maximum number of stack frames that could be present in memory is ‘n’
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