Two nodes of a BST are swapped, correct the BST | Set-2
Given a Binary Search Tree with two of the nodes of the Binary Search Tree (BST) swapped. The task is to fix (or correct) the BST.
Note: The BST will not have duplicates.
Examples:
Input Tree:
10
/ \
5 8
/ \
2 20
In the above tree, nodes 20 and 8 must be swapped to fix the tree.
Following is the output tree
10
/ \
5 20
/ \
2 8
Approach:
- Traverse the BST in In-order fashion and store the nodes in a vector.
- Then this vector is sorted after creating a copy of it.
- Use Insertion sort as it works the best and fastest when the array is almost sorted. As in this problem, only two elements will be displaced so Insertion sort here will work in linear time.
- After sorting, compare the sorted vector and the copy of the vector created earlier, by this, find out the error-some nodes and fix them by finding them in the tree and exchanging them.
Below is the implementation of the above approach:
C++
// C++ implementation of the above 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, *right; node(int x) { data = x; left = right = NULL; }}; // Utility function for insertion sortvoid insertionSort(vector<int>& v, int n){ int i, key, j; for (i = 1; i < n; i++) { key = v[i]; j = i - 1; while (j >= 0 && v[j] > key) { v[j + 1] = v[j]; j = j - 1; } v[j + 1] = key; }} // Utility function to create a vector // with inorder traversal of a binary treevoid inorder(node* root, vector<int>& v){ // Base cases if (!root) return; // Recurive call for left subtree inorder(root->left, v); // Insert node into vector v.push_back(root->data); // Recursive call for right subtree inorder(root->right, v);} // Function to exchange data // for incorrect nodesvoid find(node* root, int res, int res2){ // Base cases if (!root) { return; } // Recurive call to find // the node in left subtree find(root->left, res, res2); // Check if current node // is incorrect and exchange if (root->data == res) { root->data = res2; } else if (root->data == res2) { root->data = res; } // Recurive call to find // the node in right subtree find(root->right, res, res2);} // Primary function to fix the two nodesstruct node* correctBST(struct node* root){ // Vector to store the // inorder traversal of tree vector<int> v; // Function call to insert // nodes into vector inorder(root, v); // create a copy of the vector vector<int> v1 = v; // Sort the original vector thereby // making it a valid BST's inorder insertionSort(v, v.size()); // Traverse through both vectors and // compare misplaced values in original BST for (int i = 0; i < v.size(); i++) { // Find the mismatched values // and interchange them if (v[i] != v1[i]) { // Find and exchange the // data of the two nodes find(root, v1[i], v[i]); // As it given only two values are // wrong we don't need to check further break; } } // Return the root of corrected BST return root;}// A utility function to// print Inoder traversal void printInorder(struct node* node){ if (node == NULL) return; printInorder(node->left); printf("%d ", node->data); printInorder(node->right);} int main(){ struct node* root = new node(6); root->left = new node(10); root->right = new node(2); root->left->left = new node(1); root->left->right = new node(3); root->right->right = new node(12); root->right->left = new node(7); printf("Inorder Traversal of the"); printf("original tree \n"); printInorder(root); correctBST(root); printf("\nInorder Traversal of the"); printf("fixed tree \n"); printInorder(root); return 0;} |
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Output
Inorder Traversal of theoriginal tree 1 10 3 6 7 2 12 Inorder Traversal of thefixed tree 1 2 3 6 7 10 12
Time Complexity: O(N)
Auxiliary Space: O(N), where N is the number of nodes in the Binary Tree.
Method 2:
To understand this, you need to first understand Morris Traversal or Morris Threading Traversal. It makes use of leaf nodes’ right/left pointer to achieve O(1) space Traversal on a Binary Tree.
So, in this approach, we can solve this in O(n) time and O(1) space i.e constant space, with a single traversal of the given BST. Since the inorder traversal of BST is always a sorted array, the problem can be reduced to a problem where two elements of a sorted array are swapped.
Below is the implementation of the above approach:
C++
// C++ implementation of the// above 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, *right; }; // A utility function to swap two integers void swap( int* a, int* b ) { int t = *a; *a = *b; *b = t; } /* Helper function that allocatesa 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); }// Function for inorder traversal using // Morris Traversalvoid MorrisTraversal(struct node* root, struct node* &first, struct node* &last, struct node* &prev ) { // Current node struct node* curr = root; while(curr != NULL) { if(curr->left==NULL) { // If this node is smaller than // the previous node, it's // violating the BST rule. if(first == NULL && prev != NULL && prev->data > curr->data) { // If this is first violation, // mark these two nodes as // 'first' and 'last' first = prev; last = curr; } if(first != NULL && prev->data > curr->data) { // If this is second violation, // mark this node as last last = curr; } prev = curr; curr = curr->right; } else { /* Find the inorder predecessor of current */ struct node* pre = curr->left; while(pre->right!=NULL && pre->right!=curr) { pre = pre->right; } /* Make current as right child of its inorder predecessor */ if(pre->right==NULL) { pre->right = curr; curr = curr->left; } else { // If this node is smaller than // the previous node, it's // violating the BST rule. if(first == NULL && prev!=NULL && prev->data > curr->data) { // If this is first violation, // mark these two nodes as // 'first' and 'last' first = prev; last = curr; } if(first != NULL && prev->data > curr->data) { // If this is second violation, // mark this node as last last = curr; } prev = curr; /* Revert the changes made in the 'if' part to restore the original tree i.e., fix the right child of predecessor*/ pre->right = NULL; curr = curr->right; } } } } // A function to fix a given BST // where two nodes are swapped. This // function uses correctBSTUtil()// to find out two nodes and swaps the // nodes to fix the BST void correctBST( struct node* root ) { // Initialize pointers needed // for correctBSTUtil() struct node* first =NULL ; struct node* last = NULL ; struct node* prev =NULL ; // Set the poiters to find out two nodes MorrisTraversal( root ,first ,last , prev); // Fix (or correct) the tree swap( &(first->data), &(last->data) ); // else nodes have not been swapped, // passed tree is really BST. } /* A utility function to print Inoder traversal */void printInorder(struct node* node) { if (node == NULL) return; printInorder(node->left); printf("%d ", node->data); printInorder(node->right); } /* Driver Code */int main() { /* 6 / \ 10 2 / \ / \ 1 3 7 12 10 and 2 are swapped */ struct node *root = newNode(6); root->left = newNode(10); root->right = newNode(2); root->left->left = newNode(1); root->left->right = newNode(3); root->right->right = newNode(12); root->right->left = newNode(7); printf("Inorder Traversal of the original tree \n"); printInorder(root); correctBST(root); printf("\nInorder Traversal of the fixed tree \n"); printInorder(root); return 0; } // This code is contributed by // Sagara Jangra and Naresh Saharan |
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Java
// Java program to correct the BST // if two nodes are swapped import java.util.*; import java.lang.*; import java.io.*; class Node { int data; Node left, right; Node(int d) { data = d; left = right = null; } } class BinaryTree { Node first, last, prev; // This function does inorder traversal // Using Morris Traversal to find out the two // swapped nodes. void MorrisTraversal( Node root) { // current node Node curr = root; Node pre = null; while(curr != null) { if(curr.left==null) { // If this node is smaller than // the previous node, it's // violating the BST rule. if(first == null && prev != null && prev.data > curr.data) { // If this is first violation, // mark these two nodes as // 'first' and 'last' first = prev; last = curr; } if(first != null && prev.data > curr.data) { // If this is second violation, // mark this node as last last = curr; } prev = curr; curr = curr.right; } else { /* Find the inorder predecessor of current */ pre = curr.left; while(pre.right!=null && pre.right!=curr) { pre = pre.right; } // Make current as right child of // its inorder predecessor */ if(pre.right==null) { pre.right = curr; curr = curr.left; } else { // If this node is smaller than // the previous node, it's // violating the BST rule. if(first == null && prev!=null && prev.data > curr.data) { // If this is first violation, // mark these two nodes as // 'first' and 'last' first = prev; last = curr; } if(first != null && prev.data > curr.data) { // If this is second violation, // mark this node as last last = curr; } prev = curr; /* Revert the changes made in the 'if' part to restore the original tree i.e., fix the right child of predecessor*/ pre.right = null; curr = curr.right; } } } } // A function to fix a given BST where // two nodes are swapped. This function // uses correctBSTUtil() to find out // two nodes and swaps the nodes to // fix the BST void correctBST( Node root ) { // Initialize pointers needed // for correctBSTUtil() first = last = prev = null; // Set the poiters to find out // two nodes MorrisTraversal( root ); // Fix (or correct) the tree int temp = first.data; first.data = last.data; last.data = temp; } /* A utility function to print Inoder traversal */ void printInorder(Node node) { if (node == null) return; printInorder(node.left); System.out.print(" " + node.data); printInorder(node.right); } // Driver Code public static void main (String[] args) { /* 6 / \ 10 2 / \ / \ 1 3 7 12 10 and 2 are swapped */ Node root = new Node(6); root.left = new Node(10); root.right = new Node(2); root.left.left = new Node(1); root.left.right = new Node(3); root.right.right = new Node(12); root.right.left = new Node(7); System.out.println("Inorder Traversal"+ " of the original tree"); BinaryTree tree = new BinaryTree(); tree.printInorder(root); tree.correctBST(root); System.out.println("\nInorder Traversal"+ " of the fixed tree"); tree.printInorder(root); } } // This code is contributed by // Naresh Saharan and Sagara Jangra |
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C#
// C# program to correct the BST // if two nodes are swapped using System;public class Node { public int data; public Node left, right; public Node(int d) { data = d; left = right = null; } } public class GFG { Node first, last, prev; // This function does inorder traversal // Using Morris Traversal to find out the two // swapped nodes. void MorrisTraversal( Node root) { // current node Node curr = root; Node pre = null; while(curr != null) { if(curr.left==null) { // If this node is smaller than // the previous node, it's // violating the BST rule. if(first == null && prev != null && prev.data > curr.data) { // If this is first violation, // mark these two nodes as // 'first' and 'last' first = prev; last = curr; } if(first != null && prev.data > curr.data) { // If this is second violation, // mark this node as last last = curr; } prev = curr; curr = curr.right; } else { /* Find the inorder predecessor of current */ pre = curr.left; while(pre.right!=null && pre.right!=curr) { pre = pre.right; } // Make current as right child of // its inorder predecessor */ if(pre.right==null) { pre.right = curr; curr = curr.left; } else { // If this node is smaller than // the previous node, it's // violating the BST rule. if(first == null && prev!=null && prev.data > curr.data) { // If this is first violation, // mark these two nodes as // 'first' and 'last' first = prev; last = curr; } if(first != null && prev.data > curr.data) { // If this is second violation, // mark this node as last last = curr; } prev = curr; /* Revert the changes made in the 'if' part to restore the original tree i.e., fix the right child of predecessor*/ pre.right = null; curr = curr.right; } } } } // A function to fix a given BST where // two nodes are swapped. This function // uses correctBSTUtil() to find out // two nodes and swaps the nodes to // fix the BST void correctBST( Node root ) { // Initialize pointers needed // for correctBSTUtil() first = last = prev = null; // Set the poiters to find out // two nodes MorrisTraversal( root ); // Fix (or correct) the tree int temp = first.data; first.data = last.data; last.data = temp; } /* A utility function to print Inoder traversal */ void printInorder(Node node) { if (node == null) return; printInorder(node.left); Console.Write(" " + node.data); printInorder(node.right); } // Driver Code public static void Main(String[] args) { /* 6 / \ 10 2 / \ / \ 1 3 7 12 10 and 2 are swapped */ Node root = new Node(6); root.left = new Node(10); root.right = new Node(2); root.left.left = new Node(1); root.left.right = new Node(3); root.right.right = new Node(12); root.right.left = new Node(7); Console.WriteLine("Inorder Traversal"+ " of the original tree"); GFG tree = new GFG(); tree.printInorder(root); tree.correctBST(root); Console.WriteLine("\nInorder Traversal"+ " of the fixed tree"); tree.printInorder(root); } } // This code contributed by gauravrajput1 |
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Output
Inorder Traversal of the original tree 1 10 3 6 7 2 12 Inorder Traversal of the fixed tree 1 2 3 6 7 10 12
Time Complexity: O(N)
Auxiliary Space: O(1)
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