Two nodes of a BST are swapped, correct the BST
Two of the nodes of a Binary Search Tree (BST) are swapped. Fix (or correct) the BST.
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 8The inorder traversal of a BST produces a sorted array. So a simple method is to store inorder traversal of the input tree in an auxiliary array. Sort the auxiliary array. Finally, insert the auxiliary array elements back to the BST, keeping the structure of the BST same. The time complexity of this method is O(nLogn) and the auxiliary space needed is O(n).
We can solve this in O(n) time and with a single traversal of the given BST. Since 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. There are two cases that we need to handle:
1. The swapped nodes are not adjacent in the inorder traversal of the BST.
For example, Nodes 5 and 25 are swapped in {3 5 7 8 10 15 20 25}.
The inorder traversal of the given tree is 3 25 7 8 10 15 20 5 If we observe carefully, during inorder traversal, we find node 7 is smaller than the previous visited node 25. Here save the context of node 25 (previous node). Again, we find that node 5 is smaller than the previous node 20. This time, we save the context of node 5 (the current node ). Finally, swap the two node’s values.
2. The swapped nodes are adjacent in the inorder traversal of BST.
For example, Nodes 7 and 8 are swapped in {3 5 7 8 10 15 20 25}.
The inorder traversal of the given tree is 3 5 8 7 10 15 20 25 Unlike case #1, here only one point exists where a node value is smaller than the previous node value. e.g. node 7 is smaller than node 8.
How to Solve? We will maintain three-pointers, first, middle, and last. When we find the first point where the current node value is smaller than the previous node value, we update the first with the previous node & the middle with the current node. When we find the second point where the current node value is smaller than the previous node value, we update the last with the current node. In the case of #2, we will never find the second point. So, the last pointer will not be updated. After processing, if the last node value is null, then two swapped nodes of BST are adjacent.
Following is the implementation of the given code.
C++
// Two nodes in the BST's swapped, correct the BST.
#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 allocates a 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);
}
// This function does inorder traversal to find out the two swapped nodes.
// It sets three pointers, first, middle and last. If the swapped nodes are
// adjacent to each other, then first and middle contain the resultant nodes
// Else, first and last contain the resultant nodes
void correctBSTUtil( struct node* root, struct node** first,
struct node** middle, struct node** last,
struct node** prev )
{
if( root )
{
// Recur for the left subtree
correctBSTUtil( root->left, first, middle, last, prev );
// If this node is smaller than the previous node, it's violating
// the BST rule.
if (*prev && root->data < (*prev)->data)
{
// If this is first violation, mark these two nodes as
// 'first' and 'middle'
if ( !*first )
{
*first = *prev;
*middle = root;
}
// If this is second violation, mark this node as last
else
*last = root;
}
// Mark this node as previous
*prev = root;
// Recur for the right subtree
correctBSTUtil( root->right, first, middle, last, prev );
}
}
// 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, *middle, *last, *prev;
first = middle = last = prev = NULL;
// Set the pointers to find out two nodes
correctBSTUtil( root, &first, &middle, &last, &prev );
// Fix (or correct) the tree
if( first && last )
swap( &(first->data), &(last->data) );
else if( first && middle ) // Adjacent nodes swapped
swap( &(first->data), &(middle->data) );
// else nodes have not been swapped, passed tree is really BST.
}
/* A utility function to print Inorder traversal */
void printInorder(struct node* node)
{
if (node == NULL)
return;
printInorder(node->left);
cout <<" "<< node->data;
printInorder(node->right);
}
/* Driver program to test above functions*/
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);
cout <<"Inorder Traversal of the original tree \n";
printInorder(root);
correctBST(root);
cout <<"\nInorder Traversal of the fixed tree \n";
printInorder(root);
return 0;
}
// This code is contributed by shivanisinghss2110C
// Two nodes in the BST's swapped, correct the BST.
#include <stdio.h>
#include <stdlib.h>
/* 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 allocates a 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);
}
// This function does inorder traversal to find out the two swapped nodes.
// It sets three pointers, first, middle and last. If the swapped nodes are
// adjacent to each other, then first and middle contain the resultant nodes
// Else, first and last contain the resultant nodes
void correctBSTUtil( struct node* root, struct node** first,
struct node** middle, struct node** last,
struct node** prev )
{
if( root )
{
// Recur for the left subtree
correctBSTUtil( root->left, first, middle, last, prev );
// If this node is smaller than the previous node, it's violating
// the BST rule.
if (*prev && root->data < (*prev)->data)
{
// If this is first violation, mark these two nodes as
// 'first' and 'middle'
if ( !*first )
{
*first = *prev;
*middle = root;
}
// If this is second violation, mark this node as last
else
*last = root;
}
// Mark this node as previous
*prev = root;
// Recur for the right subtree
correctBSTUtil( root->right, first, middle, last, prev );
}
}
// 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, *middle, *last, *prev;
first = middle = last = prev = NULL;
// Set the pointers to find out two nodes
correctBSTUtil( root, &first, &middle, &last, &prev );
// Fix (or correct) the tree
if( first && last )
swap( &(first->data), &(last->data) );
else if( first && middle ) // Adjacent nodes swapped
swap( &(first->data), &(middle->data) );
// else nodes have not been swapped, passed tree is really BST.
}
/* A utility function to print Inorder traversal */
void printInorder(struct node* node)
{
if (node == NULL)
return;
printInorder(node->left);
printf("%d ", node->data);
printInorder(node->right);
}
/* Driver program to test above functions*/
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;
}
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, middle, last, prev;
// This function does inorder traversal
// to find out the two swapped nodes.
// It sets three pointers, first, middle
// and last. If the swapped nodes are
// adjacent to each other, then first
// and middle contain the resultant nodes
// Else, first and last contain the
// resultant nodes
void correctBSTUtil( Node root)
{
if( root != null )
{
// Recur for the left subtree
correctBSTUtil( root.left);
// If this node is smaller than
// the previous node, it's
// violating the BST rule.
if (prev != null && root.data <
prev.data)
{
// If this is first violation,
// mark these two nodes as
// 'first' and 'middle'
if (first == null)
{
first = prev;
middle = root;
}
// If this is second violation,
// mark this node as last
else
last = root;
}
// Mark this node as previous
prev = root;
// Recur for the right subtree
correctBSTUtil( root.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 = middle = last = prev = null;
// Set the pointers to find out
// two nodes
correctBSTUtil( root );
// Fix (or correct) the tree
if( first != null && last != null )
{
int temp = first.data;
first.data = last.data;
last.data = temp;
}
// Adjacent nodes swapped
else if( first != null && middle !=
null )
{
int temp = first.data;
first.data = middle.data;
middle.data = temp;
}
// else nodes have not been swapped,
// passed tree is really BST.
}
/* A utility function to print
Inorder traversal */
void printInorder(Node node)
{
if (node == null)
return;
printInorder(node.left);
System.out.print(" " + node.data);
printInorder(node.right);
}
// Driver program to test above functions
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 Chhavi
Python3
# Python3 program to correct the BST
# if two nodes are swapped
class Node:
# Constructor to create a new node
def __init__(self, data):
self.key = data
self.left = None
self.right = None
# Utility function to track the nodes
# that we have to swap
def correctBstUtil(root, first, middle,
last, prev):
if(root):
# Recur for the left sub tree
correctBstUtil(root.left, first,
middle, last, prev)
# If this is the first violation, mark these
# two nodes as 'first and 'middle'
if(prev[0] and root.key < prev[0].key):
if(not first[0]):
first[0] = prev[0]
middle[0] = root
else:
# If this is the second violation,
# mark this node as last
last[0] = root
prev[0] = root
# Recur for the right subtree
correctBstUtil(root.right, first,
middle, last, prev)
# 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
def correctBst(root):
# Followed four lines just for forming
# an array with only index 0 filled
# with None and we will update it accordingly.
# we made it null so that we can fill
# node data in them.
first = [None]
middle = [None]
last = [None]
prev = [None]
# Setting arrays (having zero index only)
# for capturing the required node
correctBstUtil(root, first, middle,
last, prev)
# Fixing the two nodes
if(first[0] and last[0]):
# Swapping for first and last key values
first[0].key, last[0].key = (last[0].key,
first[0].key)
elif(first[0] and middle[0]):
# Swapping for first and middle key values
first[0].key, middle[0].key = (middle[0].key,
first[0].key)
# else tree will be fine
# Function to print inorder
# traversal of tree
def PrintInorder(root):
if(root):
PrintInorder(root.left)
print(root.key, end = " ")
PrintInorder(root.right)
else:
return
# Driver code
# 6
# / \
# 10 2
# / \ / \
# 1 3 7 12
# Following 7 lines are for tree formation
root = Node(6)
root.left = Node(10)
root.right = Node(2)
root.left.left = Node(1)
root.left.right = Node(3)
root.right.left = Node(7)
root.right.right = Node(12)
# Printing inorder traversal of normal tree
print("inorder traversal of normal tree")
PrintInorder(root)
print("")
# Function call to do the task
correctBst(root)
# Printing inorder for corrected Bst tree
print("")
print("inorder for corrected BST")
PrintInorder(root)
# This code is contributed by rajutkarshai
C#
// C# program to correct the BST
// if two nodes are swapped
using System;
class Node{
public int data;
public Node left, right;
public Node(int d)
{
data = d;
left = right = null;
}
}
class BinaryTree
{
Node first, middle,
last, prev;
// This function does inorder traversal
// to find out the two swapped nodes.
// It sets three pointers, first, middle
// and last. If the swapped nodes are
// adjacent to each other, then first
// and middle contain the resultant nodes
// Else, first and last contain the
// resultant nodes
void correctBSTUtil( Node root)
{
if( root != null )
{
// Recur for the
// left subtree
correctBSTUtil(root.left);
// If this node is smaller than
// the previous node, it's
// violating the BST rule.
if (prev != null && root.data <
prev.data)
{
// If this is first violation,
// mark these two nodes as
// 'first' and 'middle'
if (first == null)
{
first = prev;
middle = root;
}
// If this is second violation,
// mark this node as last
else
last = root;
}
// Mark this node
// as previous
prev = root;
// Recur for the
// right subtree
correctBSTUtil(root.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 = middle = last =
prev = null;
// Set the pointers to
// find out two nodes
correctBSTUtil(root);
// Fix (or correct)
// the tree
if(first != null &&
last != null)
{
int temp = first.data;
first.data = last.data;
last.data = temp;
}
// Adjacent nodes swapped
else if(first != null &&
middle != null)
{
int temp = first.data;
first.data = middle.data;
middle.data = temp;
}
// else nodes have not been
// swapped, passed tree is
// really BST.
}
// A utility function to print
// Inorder 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");
BinaryTree tree = new BinaryTree();
tree.printInorder(root);
tree.correctBST(root);
Console.WriteLine("\nInorder Traversal" +
" of the fixed tree");
tree.printInorder(root);
}
}
// This code is contributed by gauravrajput1Javascript
<script>
// JavaScript program to correct the BST
// if two nodes are swapped
class Node {
constructor(val) {
this.data = val;
this.left = null;
this.right = null;
}
}
var first, middle, last, prev;
// This function does inorder traversal
// to find out the two swapped nodes.
// It sets three pointers, first, middle
// and last. If the swapped nodes are
// adjacent to each other, then first
// and middle contain the resultant nodes
// Else, first and last contain the
// resultant nodes
function correctBSTUtil(root) {
if (root != null) {
// Recur for the left subtree
correctBSTUtil(root.left);
// If this node is smaller than
// the previous node, it's
// violating the BST rule.
if (prev != null && root.data < prev.data) {
// If this is first violation,
// mark these two nodes as
// 'first' and 'middle'
if (first == null) {
first = prev;
middle = root;
}
// If this is second violation,
// mark this node as last
else
last = root;
}
// Mark this node as previous
prev = root;
// Recur for the right subtree
correctBSTUtil(root.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
function correctBST(root) {
// Initialize pointers needed
// for correctBSTUtil()
first = middle = last = prev = null;
// Set the pointers to find out
// two nodes
correctBSTUtil(root);
// Fix (or correct) the tree
if (first != null && last != null) {
var temp = first.data;
first.data = last.data;
last.data = temp;
}
// Adjacent nodes swapped
else if (first != null && middle != null) {
var temp = first.data;
first.data = middle.data;
middle.data = temp;
}
// else nodes have not been swapped,
// passed tree is really BST.
}
/*
* A utility function to print Inorder traversal
*/
function printInorder(node) {
if (node == null)
return;
printInorder(node.left);
document.write(" " + node.data);
printInorder(node.right);
}
// Driver program to test above functions
/*
* 6 / \ 10 2 / \ / \ 1 3 7 12
*
* 10 and 2 are swapped
*/
var 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);
document.write("Inorder Traversal" +
" of the original tree<br/>");
printInorder(root);
correctBST(root);
document.write("<br/>Inorder Traversal" +
" of the fixed tree<br/>");
printInorder(root);
// This code contributed by aashish1995
</script>
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)
See this for different test cases of the above code.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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