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

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

Examples:

Input:

Output: 4 

Approach: Just traverse the node from root to left recursively until left is NULL. The node whose left is NULL is the node with the minimum 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 minValue(struct node* node)
{
    if (node->left == NULL)
        return node->data;
    return minValue(node->left);
}

// 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 << minValue(root);

    return 0;
}

Java

// Java Implementation of the above 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 minValue(Node node)
{
    if (node.left == null)
        return node.data;
    return minValue(node.left);
}

// Driver code
public static void main(String args[])
{
    // Create the BST
    Node root = null;
    root = insert(root, 4);
    insert(root, 2);
    insert(root, 1);
    insert(root, 3);
    insert(root, 6);
    insert(root, 5);

    System.out.println(minValue(root));

}
}

// This code has been contributed by 29AjayKumar

Python3

# Python3 Implementation of
# the above approach
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 null left and right pointers.
def insert(node, data):
    if node is None :
        return Node(data)
    else:
        if data <= node.data:
            node.left = insert(node.left, data)
        else:
            node.right = insert(node.right, data)

        return node

# Function to return the minimum node
# in the given binary search tree
def minValue(node):
    if node.left == None:
        return node.data
    return minValue(node.left)

# Driver code
if __name__ == "__main__" :

    # Create the BST
    root = None
    root = insert(root, 4)
    insert(root, 2)
    insert(root, 1)
    insert(root, 3)
    insert(root, 6)
    insert(root, 5)

    print(minValue(root))

# This code is contributed by vinayak

C#

// C# Implementation of the above 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 minValue(Node node)
{
    if (node.left == null)
        return node.data;
    return minValue(node.left);
}

// Driver code
public static void Main(String []args)
{
    // Create the BST
    Node root = null;
    root = insert(root, 4);
    insert(root, 2);
    insert(root, 1);
    insert(root, 3);
    insert(root, 6);
    insert(root, 5);

    Console.WriteLine(minValue(root));

}
}

// This code contributed by Rajput-Ji

Javascript

<script>

// Javascript implementation of the above approach

// A binary tree node has data, pointer to
// left child and a pointer to right child
class Node
{
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right = null;
    }
};

// Helper function that allocates a new node
// with the given data and null left and right
// pointers.
function newNode(data)
{
    var 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).
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 minValue(node)
{
    if (node.left == null)
        return node.data;

    return minValue(node.left);
}

// Driver code

// Create the BST
var root = null;
root = insert(root, 4);
insert(root, 2);
insert(root, 1);
insert(root, 3);
insert(root, 6);
insert(root, 5);

document.write(minValue(root));

// This code is contributed by noob2000

</script>

Output:

1

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

Auxiliary Space : O(n), maximum number of stack frames that could be present in memory is ‘n’