Find the minimum absolute difference in two different BST's - GeeksforGeeks

21 Feb, 2022
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. 
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.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.Repeat above steps till both the BST’s are pointing to a valid nodes.
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.
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.
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.
Repeat above steps till both the BST’s are pointing to a valid nodes.
C++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Node of Binary treestruct 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 BSTvoid next(stack<node*>& it){     node* curr = it.top()->right;    it.pop();    while (curr != NULL)        it.push(curr), curr = curr->left;} // Function to find minimum differenceint 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 codeint 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;}
// Java implementation of the approachimport java.util.*; class GFG{ // Node of Binary treestatic 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 BSTstatic 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 differencestatic 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 codepublic 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
# Python3 implementation of the approachimport sys # Node of the binary treeclass node:         def __init__ (self, key):                 self.data = key        self.left = None        self.right = None # Function to iterate to the# next element of the BSTdef next(it):     curr = it[-1].right     del it[-1]    while (curr != None):        it.append(curr)        curr = curr.left             return it # Function to find minimum differencedef minDiff(root1, root2):     # Iterator for two Binary Search Trees    it1, it2 = [], []     # Initializing first iterator    curr = root1    while (curr != None):        it1.append(curr)        curr = curr.left     # Initializing second iterator    curr = root2    while (curr != None):        it2.append(curr)        curr = curr.left     # Variable to store final answer    ans = sys.maxsize     # Two pointer technique    while (len(it1) > 0 and len(it2) > 0):                 # Value it1 and it2 are pointing to        v1 = it1[-1].data        v2 = it2[-1].data         # Updating final answer        ans = min(abs(v1 - v2), ans)         # Case when v1 < v2        if (v1 < v2):            it1 = next(it1)        else:            it2 = next(it2)     # Return ans    return ans # Driver codeif __name__ == '__main__':         # BST-1    #       5    #     /   \    #    3     7    #   / \   / \    #  2   4 6   8    root2 = node(5)    root2.left = node(3)    root2.right = node(7)    root2.left.left = node(2)    root2.left.right = node(4)    root2.right.left = node(6)    root2.right.right = node(8)      # BST-2    #    11    #      \    #       15    #    root1 = node(11)    root1.right = node(15)     print(minDiff(root1, root2)) # This code is contributed by mohit kumar 29
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG{ // Node of Binary treeclass 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 BSTstatic 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 differencestatic 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 codepublic 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
<script>// Javascript implementation of the approach // Node of the binary treeclass node{    constructor(data)    {        this.data = data;        this.left = null;        this.right = null;    }} // Function to iterate to the// next element of the BSTfunction next(it){    let curr = it[it.length-1].right;    it.pop();    while (curr != null)    {        it.push(curr);        curr = curr.left;    }} // Function to find minimum differencefunction minDiff(root1,root2){    // Iterator for two Binary Search Trees    let it1 = [];    let it2 = [];      // Initializing first iterator    let 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    let ans = Number.MAX_VALUE;      // Two pointer technique    while (it1.length > 0 && it2.length > 0)    {          // value it1 and it2 are pointing to        let v1 = it1[it1.length-1].data;        let v2 = it2[it2.length-1].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// BST-1      /* 5        / \    3     7    / \ / \    2 4 6 8 */    let 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    */    let root1 = new node(11);    root1.right = new node(15);      document.write(minDiff(root1, root2)); // This code is contributed by patel2127</script>
3
 
Time complexity: O(N1 + N2) where N1 and N2 are the number 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. 
29AjayKumar
Rajput-Ji
mohit kumar 29
patel2127
sumitgumber28
Binary Search Tree
Data Structures
Tree
Data Structures
Binary Search Tree
Tree
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