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:
BST_LCA
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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Output:

1

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



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