Find the minimum absolute difference in two different BST’s

Given 2 Binary Search Trees, select one node from each tree such that their absolute difference is minimum possible. Assume each BST has at-least one node.

Examples:

Input : N1 = 7, N2 = 2 

BST1 :
          5 
        /   \ 
       3     7 
      / \   / \ 
     2   4 6   8
     
BST2 :
         11
           \
            13
   
Output : 3
8 is largest number in the first BST
and 11 is smallest in the second.
Thus, the final answer will be 11-8 = 3


Input : N1 = 4, N2 = 2
BST1 :
          3 
        /   \ 
       2     4
              \
              14
        
BST2 :
          7
           \
            13
   
Output : 1

Approach:
The idea is to use the two-pointer technique and iterating the pointers using the following steps.



  1. Create forward iterators for both the BST’s. Let’s say that the value of nodes they are pointing at are v1 and v2 respectively.
  2. Now at each step:
    • Update final ans as min(ans, abs(v1-v2)) .
    • If v1 < v2, move iterator of first BST else move the iterator of the second BST.
  3. Repeat above steps till both the BST’s are pointing to a valid nodes.

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Node of Binary tree
struct node {
    int data;
    node* left;
    node* right;
    node(int data)
    {
        this->data = data;
        left = NULL;
        right = NULL;
    }
};
  
// Function to iterate to the
// next element of the BST
void next(stack<node*>& it)
{
  
    node* curr = it.top()->right;
    it.pop();
    while (curr != NULL)
        it.push(curr), curr = curr->left;
}
  
// Function to find minimum difference
int minDiff(node* root1, node* root2)
{
  
    // Iterator for two Binary Search Trees
    stack<node *> it1, it2;
  
    // Initializing first iterator
    node* curr = root1;
    while (curr != NULL)
        it1.push(curr), curr = curr->left;
  
    // Initializing second iterator
    curr = root2;
    while (curr != NULL)
        it2.push(curr), curr = curr->left;
  
    // Variable to store final answer
    int ans = INT_MAX;
  
    // Two pointer technique
    while (it1.size() and it2.size()) {
  
        // value it1 and it2 are pointing to
        int v1 = it1.top()->data;
        int v2 = it2.top()->data;
  
        // Updating final answer
        ans = min(abs(v1 - v2), ans);
  
        // Case when v1 < v2
        if (v1 < v2)
            next(it1);
        else
            next(it2);
    }
  
    // Return ans
    return ans;
}
  
// Driver code
int main()
{
    // BST-1
  
    /*    5 
        /   \ 
       3     7 
      / \   / \ 
     2   4 6   8 */
    node* root2 = new node(5);
    root2->left = new node(3);
    root2->right = new node(7);
    root2->left->left = new node(2);
    root2->left->right = new node(4);
    root2->right->left = new node(6);
    root2->right->right = new node(8);
  
    // BST-2
  
    /*  11
         \
          15
    */
    node* root1 = new node(11);
    root1->right = new node(15);
  
    cout << minDiff(root1, root2);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Node of Binary tree
static class node 
{
    int data;
    node left;
    node right;
    node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
};
  
// Function to iterate to the
// next element of the BST
static void next(Stack<node> it)
{
    node curr = it.peek().right;
    it.pop();
    while (curr != null)
    {
        it.push(curr);
        curr = curr.left;
    }
}
  
// Function to find minimum difference
static int minDiff(node root1, node root2)
{
  
    // Iterator for two Binary Search Trees
    Stack<node> it1 = new Stack<node>();
    Stack<node> it2 = new Stack<node>();
  
    // Initializing first iterator
    node curr = root1;
    while (curr != null)
    {
        it1.push(curr);
        curr = curr.left;
    }
  
    // Initializing second iterator
    curr = root2;
    while (curr != null)
    {
        it2.push(curr);
        curr = curr.left;
    }
  
    // Variable to store final answer
    int ans = Integer.MAX_VALUE;
  
    // Two pointer technique
    while (it1.size() > 0 && it2.size() > 0
    {
  
        // value it1 and it2 are pointing to
        int v1 = it1.peek().data;
        int v2 = it2.peek().data;
  
        // Updating final answer
        ans = Math.min(Math.abs(v1 - v2), ans);
  
        // Case when v1 < v2
        if (v1 < v2)
            next(it1);
        else
            next(it2);
    }
  
    // Return ans
    return ans;
}
  
// Driver code
public static void main(String[] args) 
{
    // BST-1
  
    /* 5 
        / \ 
    3     7 
    / \ / \ 
    2 4 6 8 */
    node root2 = new node(5);
    root2.left = new node(3);
    root2.right = new node(7);
    root2.left.left = new node(2);
    root2.left.right = new node(4);
    root2.right.left = new node(6);
    root2.right.right = new node(8);
  
    // BST-2
  
    /* 11
        \
        15
    */
    node root1 = new node(11);
    root1.right = new node(15);
  
    System.out.println(minDiff(root1, root2));
}
}
  
// This code is contributed by 29AjayKumar

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG
{
  
// Node of Binary tree
class node 
{
    public int data;
    public node left;
    public node right;
    public node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
};
  
// Function to iterate to the
// next element of the BST
static void next(Stack<node> it)
{
    node curr = it.Peek().right;
    it.Pop();
    while (curr != null)
    {
        it.Push(curr);
        curr = curr.left;
    }
}
  
// Function to find minimum difference
static int minDiff(node root1, node root2)
{
  
    // Iterator for two Binary Search Trees
    Stack<node> it1 = new Stack<node>();
    Stack<node> it2 = new Stack<node>();
  
    // Initializing first iterator
    node curr = root1;
    while (curr != null)
    {
        it1.Push(curr);
        curr = curr.left;
    }
  
    // Initializing second iterator
    curr = root2;
    while (curr != null)
    {
        it2.Push(curr);
        curr = curr.left;
    }
  
    // Variable to store readonly answer
    int ans = int.MaxValue;
  
    // Two pointer technique
    while (it1.Count > 0 && it2.Count > 0) 
    {
  
        // value it1 and it2 are pointing to
        int v1 = it1.Peek().data;
        int v2 = it2.Peek().data;
  
        // Updating readonly answer
        ans = Math.Min(Math.Abs(v1 - v2), ans);
  
        // Case when v1 < v2
        if (v1 < v2)
            next(it1);
        else
            next(it2);
    }
  
    // Return ans
    return ans;
}
  
// Driver code
public static void Main(String[] args) 
{
    // BST-1
  
    /* 5 
        / \ 
    3     7 
    / \ / \ 
    2 4 6 8 */
    node root2 = new node(5);
    root2.left = new node(3);
    root2.right = new node(7);
    root2.left.left = new node(2);
    root2.left.right = new node(4);
    root2.right.left = new node(6);
    root2.right.right = new node(8);
  
    // BST-2
  
    /* 11
        \
        15
    */
    node root1 = new node(11);
    root1.right = new node(15);
  
    Console.WriteLine(minDiff(root1, root2));
}
}
  
// This code is contributed by Rajput-Ji

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Output:

3

Time complexity: O(N1 + N2) where N1 and N2 are the bumber of nodes of first and second BST respectively.
Space Complexity: O(H1 + H2) where H1 and H2 are the heights of first and second BST respectively.



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