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.
C++
// 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; } |
Java
// 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 |
Python3
# Python3 implementation of the approach import sys # Node of the binary tree class node: def __init__ ( self , key): self .data = key self .left = None self .right = None # Function to iterate to the # next element of the BST def next (it): curr = it[ - 1 ].right del it[ - 1 ] while (curr ! = None ): it.append(curr) curr = curr.left return it # Function to find minimum difference def 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 code if __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#
// 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 |
Javascript
<script> // Javascript implementation of the approach // Node of the binary tree class node { constructor(data) { this .data = data; this .left = null ; this .right = null ; } } // Function to iterate to the // next element of the BST function next(it) { let curr = it[it.length-1].right; it.pop(); while (curr != null ) { it.push(curr); curr = curr.left; } } // Function to find minimum difference function 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.
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