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Maximum element between two nodes of BST

Given an array of N elements and two integers A, B which belongs to the given array. Create a Binary Search Tree by inserting element from arr[0] to arr[n-1]. The task is to find the maximum element in the path from A to B.

Examples :

Input : arr[] = { 18, 36, 9, 6, 12, 10, 1, 8 }, 
        a = 1, 
        b = 10.
Output : 12


Path from 1 to 10 contains { 1, 6, 9, 12, 10 }. Maximum element is 12.



The idea is to find Lowest Common Ancestor of node ‘a’ and node ‘b’. Then search maximum node between LCA and ‘a’, also find maximum node between LCA and ‘b’. Answer will be maximum node of two.

C++

// C++ program to find maximum element in the path
// between two Nodes of Binary Search Tree.
#include <bits/stdc++.h>
using namespace std;
  
struct Node
{
    struct Node *left, *right;
    int data;
};
  
// Create and return a pointer of new Node.
Node *createNode(int x)
{
    Node *p = new Node;
    p -> data = x;
    p -> left = p -> right = NULL;
    return p;
}
  
// Insert a new Node in Binary Search Tree.
void insertNode(struct Node *root, int x)
{
    Node *p = root, *q = NULL;
  
    while (p != NULL)
    {
        q = p;
        if (p -> data < x)
            p = p -> right;
        else
            p = p -> left;
    }
  
    if (q == NULL)
        p = createNode(x);
    else
    {
        if (q -> data < x)
            q -> right = createNode(x);
        else
            q -> left = createNode(x);
    }
}
  
// Return the maximum element between a Node
// and its given ancestor.
int maxelpath(Node *q, int x)
{
    Node *p = q;
  
    int mx = INT_MIN;
  
    // Traversing the path between ansector and
    // Node and finding maximum element.
    while (p -> data != x)
    {
        if (p -> data > x)
        {
            mx = max(mx, p -> data);
            p = p -> left;
        }
        else
        {
            mx = max(mx, p -> data);
            p = p -> right;
        }
    }
  
    return max(mx, x);
}
  
// Return maximum element in the path between
// two given Node of BST.
int maximumElement(struct Node *root, int x, int y)
{
    Node *p = root;
  
    // Finding the LCA of Node x and Node y
    while ((x < p -> data && y < p -> data) ||
        (x > p -> data && y > p -> data))
    {
        // Checking if both the Node lie on the
        // left side of the parent p.
        if (x < p -> data && y < p -> data)
            p = p -> left;
  
        // Checking if both the Node lie on the
        // right side of the parent p.
        else if (x > p -> data && y > p -> data)
            p = p -> right;
    }
  
    // Return the maximum of maximum elements occur
    // in path from ancestor to both Node.
    return max(maxelpath(p, x), maxelpath(p, y));
}
  
  
// Driver Code
int main()
{
    int arr[] = { 18, 36, 9, 6, 12, 10, 1, 8 };
    int a = 1, b = 10;
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Creating the root of Binary Search Tree
    struct Node *root = createNode(arr[0]);
  
    // Inserting Nodes in Binary Search Tree
    for (int i = 1; i < n; i++)
        insertNode(root, arr[i]);
  
    cout << maximumElement(root, a, b) << endl;
  
    return 0;
}

Java

// Java program to find maximum element in the path
// between two Nodes of Binary Search Tree.
class Solution
{
      
static class Node
{
     Node left, right;
    int data;
}
   
// Create and return a pointer of new Node.
static Node createNode(int x)
{
    Node p = new Node();
    p . data = x;
    p . left = p . right = null;
    return p;
}
   
// Insert a new Node in Binary Search Tree.
static void insertNode( Node root, int x)
{
    Node p = root, q = null;
   
    while (p != null)
    {
        q = p;
        if (p . data < x)
            p = p . right;
        else
            p = p . left;
    }
   
    if (q == null)
        p = createNode(x);
    else
    {
        if (q . data < x)
            q . right = createNode(x);
        else
            q . left = createNode(x);
    }
}
   
// Return the maximum element between a Node
// and its given ancestor.
static int maxelpath(Node q, int x)
{
    Node p = q;
   
    int mx = -1;
   
    // Traversing the path between ansector and
    // Node and finding maximum element.
    while (p . data != x)
    {
        if (p . data > x)
        {
            mx = Math.max(mx, p . data);
            p = p . left;
        }
        else
        {
            mx = Math.max(mx, p . data);
            p = p . right;
        }
    }
   
    return Math.max(mx, x);
}
   
// Return maximum element in the path between
// two given Node of BST.
static int maximumElement( Node root, int x, int y)
{
    Node p = root;
   
    // Finding the LCA of Node x and Node y
    while ((x < p . data && y < p . data) ||
        (x > p . data && y > p . data))
    {
        // Checking if both the Node lie on the
        // left side of the parent p.
        if (x < p . data && y < p . data)
            p = p . left;
   
        // Checking if both the Node lie on the
        // right side of the parent p.
        else if (x > p . data && y > p . data)
            p = p . right;
    }
   
    // Return the maximum of maximum elements occur
    // in path from ancestor to both Node.
    return Math.max(maxelpath(p, x), maxelpath(p, y));
}
   
   
// Driver Code
public static void main(String args[])
{
    int arr[] = { 18, 36, 9, 6, 12, 10, 1, 8 };
    int a = 1, b = 10;
    int n =arr.length;
   
    // Creating the root of Binary Search Tree
     Node root = createNode(arr[0]);
   
    // Inserting Nodes in Binary Search Tree
    for (int i = 1; i < n; i++)
        insertNode(root, arr[i]);
   
    System.out.println( maximumElement(root, a, b) );
   
}
}
//contributed by Arnab Kundu 

Python3

# Python 3 program to find maximum element
# in the path between two Nodes of Binary
# Search Tree.

# Create and return a pointer of new Node.
class createNode:

# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None

# Insert a new Node in Binary Search Tree.
def insertNode(root, x):
p, q = root, None

while p != None:
q = p
if p.data < x: p = p.right else: p = p.left if q == None: p = createNode(x) else: if q.data < x: q.right = createNode(x) else: q.left = createNode(x) # Return the maximum element between a # Node and its given ancestor. def maxelpath(q, x): p = q mx = -999999999999 # Traversing the path between ansector # and Node and finding maximum element. while p.data != x: if p.data > x:
mx = max(mx, p.data)
p = p.left
else:
mx = max(mx, p.data)
p = p.right

return max(mx, x)

# Return maximum element in the path
# between two given Node of BST.
def maximumElement(root, x, y):
p = root

# Finding the LCA of Node x and Node y
while ((x < p.data and y < p.data) or (x > p.data and y > p.data)):

# Checking if both the Node lie on
# the left side of the parent p.
if x < p.data and y < p.data: p = p.left # Checking if both the Node lie on # the right side of the parent p. elif x > p.data and y > p.data:
p = p.right

# Return the maximum of maximum elements
# occur in path from ancestor to both Node.
return max(maxelpath(p, x), maxelpath(p, y))

# Driver Code
if __name__ == ‘__main__’:
arr = [ 18, 36, 9, 6, 12, 10, 1, 8]
a, b = 1, 10
n = len(arr)

# Creating the root of Binary Search Tree
root = createNode(arr[0])

# Inserting Nodes in Binary Search Tree
for i in range(1,n):
insertNode(root, arr[i])

print(maximumElement(root, a, b))

# This code is contributed by PranchalK

C#

using System;
  
// C# program to find maximum element in the path 
// between two Nodes of Binary Search Tree. 
public class Solution
{
  
public class Node
{
     public Node left, right;
    public int data;
}
  
// Create and return a pointer of new Node. 
public static Node createNode(int x)
{
    Node p = new Node();
    p.data = x;
    p.left = p.right = null;
    return p;
}
  
// Insert a new Node in Binary Search Tree. 
public static void insertNode(Node root, int x)
{
    Node p = root, q = null;
  
    while (p != null)
    {
        q = p;
        if (p.data < x)
        {
            p = p.right;
        }
        else
        {
            p = p.left;
        }
    }
  
    if (q == null)
    {
        p = createNode(x);
    }
    else
    {
        if (q.data < x)
        {
            q.right = createNode(x);
        }
        else
        {
            q.left = createNode(x);
        }
    }
}
  
// Return the maximum element between a Node 
// and its given ancestor. 
public static int maxelpath(Node q, int x)
{
    Node p = q;
  
    int mx = -1;
  
    // Traversing the path between ansector and 
    // Node and finding maximum element. 
    while (p.data != x)
    {
        if (p.data > x)
        {
            mx = Math.Max(mx, p.data);
            p = p.left;
        }
        else
        {
            mx = Math.Max(mx, p.data);
            p = p.right;
        }
    }
  
    return Math.Max(mx, x);
}
  
// Return maximum element in the path between 
// two given Node of BST. 
public static int maximumElement(Node root, int x, int y)
{
    Node p = root;
  
    // Finding the LCA of Node x and Node y 
    while ((x < p.data && y < p.data) || (x > p.data && y > p.data))
    {
        // Checking if both the Node lie on the 
        // left side of the parent p. 
        if (x < p.data && y < p.data)
        {
            p = p.left;
        }
  
        // Checking if both the Node lie on the 
        // right side of the parent p. 
        else if (x > p.data && y > p.data)
        {
            p = p.right;
        }
    }
  
    // Return the maximum of maximum elements occur 
    // in path from ancestor to both Node. 
    return Math.Max(maxelpath(p, x), maxelpath(p, y));
}
  
  
// Driver Code 
public static void Main(string[] args)
{
    int[] arr = new int[] {18, 36, 9, 6, 12, 10, 1, 8};
    int a = 1, b = 10;
    int n = arr.Length;
  
    // Creating the root of Binary Search Tree 
     Node root = createNode(arr[0]);
  
    // Inserting Nodes in Binary Search Tree 
    for (int i = 1; i < n; i++)
    {
        insertNode(root, arr[i]);
    }
  
    Console.WriteLine(maximumElement(root, a, b));
  
}
}
  
  //  This code is contributed by Shrikant13


Output :

12

Time complexity : O(h) where h is height of BST

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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