# Convert Binary Tree to Binary Search Tree

Harshit Jindal Oct 12, 2023

A Binary Tree is a non-linear data structure. It is called a binary tree because each node has a maximum of two children. These children are called left children and right children. It can also be interpreted as an undirected graph in which the topmost node is called root. A Binary Search Tree (BST) is a binary tree with special properties that helps to keep the data organized in a sorted fashion.

In this tutorial, we will discuss how to convert a binary tree to BST while maintaining the Binary Tree’s original structure.

## Convert Binary Tree to BST Illustration ## Convert Binary Tree to BST Implementation

``````#include <bits/stdc++.h>
using namespace std;

class Node {
public:
int data;
Node *left, *right;
Node(int x) {
this->data = x;
this->left = this->right = NULL;
}
};
vector<int> v;

void inorder(Node* root) {
if (root != NULL) {
inorder(root->left);
cout << root->data << " ";
inorder(root->right);
}
}

void storetree(Node* root, int i = 0) {
if (!root) {
return;
}
storetree(root->left);
v.push_back(root->data);
storetree(root->right);
}

void restoretree(Node* root, int& i) {
if (!root) {
return;
}
restoretree(root->left, i);
root->data = v[i];
i++;
restoretree(root->right, i);
}

void converttoBST(Node* root) {
if (!root) {
return;
}
storetree(root);
sort(v.begin(), v.end());
int i = 0;
restoretree(root, i);
}

int main() {
Node* root = new Node(3);
root->left = new Node(1);
root->right = new Node(7);
root->left->left = new Node(4);
root->left->right = new Node(5);
root->left->left->right = new Node(2);
root->right->left = new Node(6);
root->right->right = new Node(9);
root->right->right->left = new Node(8);
cout << "The inorder traversal of the tree is : ";
inorder(root);
cout << endl;
converttoBST(root);
cout << "The inorder traversal of the tree is : ";
inorder(root);
cout << endl;
}
``````

## Convert Binary Tree to BST Algorithm Complexity

### Time Complexity

• Average Case

The time complexity of doing the inorder traversal in which we store the array in `sorted` and store the `sorted` array back to the binary tree is `O(n)`. But, the complexity of sorting the array is `O(nlogn)` and hence the total complexity is given as `O(nlogn) + 2*O(n)`. The time complexity is of the order of `O(nlogn)`.

• Best Case

The best-case time complexity is of the order of `O(n)`. When the given binary tree is already a BST, we do the inorder traversal to realize it, and no sorting operations are required.

• Worst Case

The worst-case time complexity is of the order of `O(nlogn)`.

### Space Complexity

The algorithm’s space complexity is `O(n)` due to the extra space required by recursion calls.

Harshit Jindal has done his Bachelors in Computer Science Engineering(2021) from DTU. He has always been a problem solver and now turned that into his profession. Currently working at M365 Cloud Security team(Torus) on Cloud Security Services and Datacenter Buildout Automation.