Given a Binary Tree, convert it to a Binary Search Tree. The conversion must be done in such a way that keeps the original structure of the Binary Tree.
This solution will use Sets of C++ STL instead of array-based solution.
Examples:
Example 1 Input: 10 / \ 2 7 / \ 8 4 Output: 8 / \ 4 10 / \ 2 7 Example 2 Input: 10 / \ 30 15 / \ 20 5 Output: 15 / \ 10 20 / \ 5 30
Solution
- Copy the items of binary tree in a set while doing inorder traversal. This takes O(n log n) time. Note that set in C++ STL is implemented using a Self Balancing Binary Search Tree like Red Black Tree, AVL Tree, etc
- There is no need to sort the set as sets in C++ are implemented using Self-balancing binary search trees due to which each operation such as insertion, searching, deletion, etc takes O(log n) time.
- Now simply copy the items of set one by one from beginning to the tree while doing inorder traversal of the tree. Care should be taken as when copying each item of set from its beginning, we first copy it to the tree while doing inorder traversal, then remove it from the set as well.
Now the above solution is simpler and easier to implement than the array-based conversion of Binary tree to Binary search tree explained here- Conversion of Binary Tree to Binary Search tree (Set-1), where we had to separately make a function to sort the items of the array after copying the items from tree to it.
Implementation: Program to convert a binary tree to a binary search tree using the set.
/* CPP program to convert a Binary tree to BST using sets as containers. */
#include <bits/stdc++.h> using namespace std;
struct Node {
int data;
struct Node *left, *right;
}; // function to store the nodes in set while // doing inorder traversal. void storeinorderInSet(Node* root, set< int >& s)
{ if (!root)
return ;
// visit the left subtree first
storeinorderInSet(root->left, s);
// insertion takes order of O(logn) for sets
s.insert(root->data);
// visit the right subtree
storeinorderInSet(root->right, s);
} // Time complexity = O(nlogn)
// function to copy items of set one by one // to the tree while doing inorder traversal void setToBST(set< int >& s, Node* root)
{ // base condition
if (!root)
return ;
// first move to the left subtree and
// update items
setToBST(s, root->left);
// iterator initially pointing to the
// beginning of set
auto it = s.begin();
// copying the item at beginning of
// set(sorted) to the tree.
root->data = *it;
// now erasing the beginning item from set.
s.erase(it);
// now move to right subtree and update items
setToBST(s, root->right);
} // T(n) = O(nlogn) time
// Converts Binary tree to BST. void binaryTreeToBST(Node* root)
{ set< int > s;
// populating the set with the tree's
// inorder traversal data
storeinorderInSet(root, s);
// now sets are by default sorted as
// they are implemented using self-
// balancing BST
// copying items from set to the tree
// while inorder traversal which makes a BST
setToBST(s, root);
} // Time complexity = O(nlogn),
// Auxiliary Space = O(n) for set.
// helper function to create a node Node* newNode( int data)
{ // dynamically allocating memory
Node* temp = new Node();
temp->data = data;
temp->left = temp->right = NULL;
return temp;
} // function to do inorder traversal void inorder(Node* root)
{ if (!root)
return ;
inorder(root->left);
cout << root->data << " " ;
inorder(root->right);
} int main()
{ Node* root = newNode(5);
root->left = newNode(7);
root->right = newNode(9);
root->right->left = newNode(10);
root->left->left = newNode(1);
root->left->right = newNode(6);
root->right->right = newNode(11);
/* Constructing tree given in the above figure
5
/ \
7 9
/\ / \
1 6 10 11 */
// converting the above Binary tree to BST
binaryTreeToBST(root);
cout << "Inorder traversal of BST is: " << endl;
inorder(root);
return 0;
} |
/* Java program to convert a Binary tree to BST using sets as containers. */ import java.util.*;
class Solution
{ static class Node
{ int data;
Node left, right;
} // set static Set<Integer> s = new HashSet<Integer>();
// function to store the nodes in set while // doing inorder traversal. static void storeinorderInSet(Node root)
{ if (root == null )
return ;
// visit the left subtree first
storeinorderInSet(root.left);
// insertion takes order of O(logn) for sets
s.add(root.data);
// visit the right subtree
storeinorderInSet(root.right);
} // Time complexity = O(nlogn)
// function to copy items of set one by one // to the tree while doing inorder traversal static void setToBST( Node root)
{ // base condition
if (root == null )
return ;
// first move to the left subtree and
// update items
setToBST( root.left);
// iterator initially pointing to the
// beginning of set
// copying the item at beginning of
// set(sorted) to the tree.
root.data = s.iterator().next();
// now erasing the beginning item from set.
s.remove(root.data);
// now move to right subtree and update items
setToBST( root.right);
} // T(n) = O(nlogn) time
// Converts Binary tree to BST. static void binaryTreeToBST(Node root)
{ s.clear();
// populating the set with the tree's
// inorder traversal data
storeinorderInSet(root);
// now sets are by default sorted as
// they are implemented using self-
// balancing BST
// copying items from set to the tree
// while inorder traversal which makes a BST
setToBST( root);
} // Time complexity = O(nlogn),
// Auxiliary Space = O(n) for set. // helper function to create a node static Node newNode( int data)
{ // dynamically allocating memory
Node temp = new Node();
temp.data = data;
temp.left = temp.right = null ;
return temp;
} // function to do inorder traversal static void inorder(Node root)
{ if (root == null )
return ;
inorder(root.left);
System.out.print(root.data + " " );
inorder(root.right);
} // Driver code public static void main(String args[])
{ Node root = newNode( 5 );
root.left = newNode( 7 );
root.right = newNode( 9 );
root.right.left = newNode( 10 );
root.left.left = newNode( 1 );
root.left.right = newNode( 6 );
root.right.right = newNode( 11 );
/* Constructing tree given in the above figure
5
/ \
7 9
/\ / \
1 6 10 11 */
// converting the above Binary tree to BST
binaryTreeToBST(root);
System.out.println( "Inorder traversal of BST is: " );
inorder(root);
} } // This code is contributed by Arnab Kundu |
# Python3 program to convert a Binary tree # to BST using sets as containers. # Binary Tree Node """ A utility function to create a new BST node """ class newNode:
# Construct to create a newNode
def __init__( self , data):
self .data = data
self .left = None
self .right = None
# function to store the nodes in set # while doing inorder traversal. def storeinorderInSet(root, s):
if ( not root) :
return
# visit the left subtree first
storeinorderInSet(root.left, s)
# insertion takes order of O(logn)
# for sets
s.add(root.data)
# visit the right subtree
storeinorderInSet(root.right, s)
# Time complexity = O(nlogn) # function to copy items of set one by one # to the tree while doing inorder traversal def setToBST(s, root) :
# base condition
if ( not root):
return
# first move to the left subtree and
# update items
setToBST(s, root.left)
# iterator initially pointing to
# the beginning of set
it = next ( iter (s))
# copying the item at beginning of
# set(sorted) to the tree.
root.data = it
# now erasing the beginning item from set.
s.remove(it)
# now move to right subtree
# and update items
setToBST(s, root.right)
# T(n) = O(nlogn) time # Converts Binary tree to BST. def binaryTreeToBST(root):
s = set ()
# populating the set with the tree's
# inorder traversal data
storeinorderInSet(root, s)
# now sets are by default sorted as
# they are implemented using self-
# balancing BST
# copying items from set to the tree
# while inorder traversal which makes a BST
setToBST(s, root)
# Time complexity = O(nlogn), # Auxiliary Space = O(n) for set. # function to do inorder traversal def inorder(root) :
if ( not root) :
return
inorder(root.left)
print (root.data, end = " " )
inorder(root.right)
# Driver Code if __name__ = = '__main__' :
root = newNode( 5 )
root.left = newNode( 7 )
root.right = newNode( 9 )
root.right.left = newNode( 10 )
root.left.left = newNode( 1 )
root.left.right = newNode( 6 )
root.right.right = newNode( 11 )
""" Constructing tree given in
the above figure
5
/ \
7 9
/\ / \
1 6 10 11 """
# converting the above Binary tree to BST
binaryTreeToBST(root)
print ( "Inorder traversal of BST is: " )
inorder(root)
# This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
// C# program to convert // a Binary tree to BST using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
class Solution{
class Node
{ public int data;
public Node left,
right;
} // set static SortedSet< int > s =
new SortedSet< int >();
// function to store the nodes // in set while doing inorder // traversal. static void storeinorderInSet(Node root)
{ if (root == null )
return ;
// visit the left subtree
// first
storeinorderInSet(root.left);
// insertion takes order of
// O(logn) for sets
s.Add(root.data);
// visit the right subtree
storeinorderInSet(root.right);
} // Time complexity = O(nlogn) // function to copy items of // set one by one to the tree // while doing inorder traversal static void setToBST(Node root)
{ // base condition
if (root == null )
return ;
// first move to the left
// subtree and update items
setToBST(root.left);
// iterator initially pointing
// to the beginning of set copying
// the item at beginning of set(sorted)
// to the tree.
root.data = s.First();
// now erasing the beginning item
// from set.
s.Remove(s.First());
// now move to right subtree and
// update items
setToBST( root.right);
} // T(n) = O(nlogn) time // Converts Binary tree to BST.
static void binaryTreeToBST(Node root)
{ s.Clear();
// populating the set with
// the tree's inorder traversal
// data
storeinorderInSet(root);
// now sets are by default sorted
// as they are implemented using
// self-balancing BST
// copying items from set to the
// tree while inorder traversal
// which makes a BST
setToBST( root);
} // Time complexity = O(nlogn), // Auxiliary Space = O(n) for set. // helper function to create a node static Node newNode( int data)
{ // dynamically allocating
// memory
Node temp = new Node();
temp.data = data;
temp.left = temp.right = null ;
return temp;
} // function to do inorder traversal static void inorder(Node root)
{ if (root == null )
return ;
inorder(root.left);
Console.Write(root.data + " " );
inorder(root.right);
} // Driver code public static void Main( string []args)
{ Node root = newNode(5);
root.left = newNode(7);
root.right = newNode(9);
root.right.left = newNode(10);
root.left.left = newNode(1);
root.left.right = newNode(6);
root.right.right = newNode(11);
/* Constructing tree given in
// the above figure
5
/ \
7 9
/\ / \
1 6 10 11 */
// converting the above Binary
// tree to BST
binaryTreeToBST(root);
Console.Write( "Inorder traversal of " +
"BST is: \n" );
inorder(root);
} } // This code is contributed by Rutvik_56 |
<script> // JavaScript program to convert // a Binary tree to BST class Node { constructor()
{
this .data = 0;
this .right = null ;
this .left = null ;
}
} // set var s = new Set();
// function to store the nodes // in set while doing inorder // traversal. function storeinorderInSet(root)
{ if (root == null )
return ;
// visit the left subtree
// first
storeinorderInSet(root.left);
// insertion takes order of
// O(logn) for sets
s.add(root.data);
// visit the right subtree
storeinorderInSet(root.right);
} // Time complexity = O(nlogn) // function to copy items of // set one by one to the tree // while doing inorder traversal function setToBST(root)
{ // base condition
if (root == null )
return ;
// first move to the left
// subtree and update items
setToBST(root.left);
var tmp = [...s].sort((a,b)=> a-b);
// iterator initially pointing
// to the beginning of set copying
// the item at beginning of set(sorted)
// to the tree.
root.data = tmp[0];
// now erasing the beginning item
// from set.
s. delete (tmp[0]);
// now move to right subtree and
// update items
setToBST( root.right);
} // T(n) = O(nlogn) time // Converts Binary tree to BST.
function binaryTreeToBST(root)
{ s = new Set();
// populating the set with
// the tree's inorder traversal
// data
storeinorderInSet(root);
// now sets are by default sorted
// as they are implemented using
// self-balancing BST
// copying items from set to the
// tree while inorder traversal
// which makes a BST
setToBST( root);
} // Time complexity = O(nlogn), // Auxiliary Space = O(n) for set. // helper function to create a node function newNode(data)
{ // dynamically allocating
// memory
var temp = new Node();
temp.data = data;
temp.left = temp.right = null ;
return temp;
} // function to do inorder traversal function inorder(root)
{ if (root == null )
return ;
inorder(root.left);
document.write(root.data + " " );
inorder(root.right);
} // Driver code var root = newNode(5);
root.left = newNode(7); root.right = newNode(9); root.right.left = newNode(10); root.left.left = newNode(1); root.left.right = newNode(6); root.right.right = newNode(11); /* Constructing tree given in // the above figure 5
/ \
7 9
/\ / \
1 6 10 11 */
// converting the above Binary // tree to BST binaryTreeToBST(root); document.write( "Inorder traversal of " +
"BST is: <br>" );
inorder(root); </script> |
Inorder traversal of BST is: 1 5 6 7 9 10 11
Complexity Analysis:
- Time Complexity: O(n Log n)
- Auxiliary Space: (n)