# Find the node with maximum value in a Binary Search Tree using recursion

• Difficulty Level : Easy
• Last Updated : 29 Jun, 2021

Given a Binary Search Tree, the task is to find the node with maximum value.

Examples:

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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 ``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

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Output:

`6`

Time Complexity: O(n), worst case happens for right skewed trees.

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