Skip to content
Related Articles

Related Articles

Binary Tree to Binary Search Tree Conversion using STL set
  • Difficulty Level : Easy
  • Last Updated : 21 Nov, 2020

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 

  1. 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
  2. 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.
  3. 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.
Program to convert a binary tree to a binary search tree using the set.
 

C++




/* 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




/* 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




# 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#




// 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


Output: 

Inorder traversal of BST is: 
1 5 6 7 9 10 11




 

Time Complexity: O(n Log n) 
Auxiliary Space: (n)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :