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# Binary Tree to Binary Search Tree Conversion

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 Binary Tree.

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:

Following is a 3 step solution for converting Binary tree to Binary Search Tree.

1. Create a temp array arr[] that stores inorder traversal of the tree. This step takes O(n) time.
2. Sort the temp array arr[]. Time complexity of this step depends upon the sorting algorithm. In the following implementation, Quick Sort is used which takes (n^2) time. This can be done in O(nLogn) time using Heap Sort or Merge Sort.
3. Again do inorder traversal of tree and copy array elements to tree nodes one by one. This step takes O(n) time.

Following is the implementation of the above approach. The main function to convert is highlighted in the following code.

Implementation:

## C++

 /* A program to convert Binary Tree to Binary Search Tree */ #include using namespace std;   /* A binary tree node structure */ struct node {     int data;     struct node* left;     struct node* right; };   /* A helper function that stores inorder    traversal of a tree rooted with node */ void storeInorder(struct node* node, int inorder[], int* index_ptr) {     // Base Case     if (node == NULL)         return;       /* first store the left subtree */     storeInorder(node->left, inorder, index_ptr);       /* Copy the root's data */     inorder[*index_ptr] = node->data;     (*index_ptr)++; // increase index for next entry       /* finally store the right subtree */     storeInorder(node->right, inorder, index_ptr); }   /* A helper function to count nodes in a Binary Tree */ int countNodes(struct node* root) {     if (root == NULL)         return 0;     return countNodes(root->left) + countNodes(root->right) + 1; }   // Following function is needed for library function qsort() int compare(const void* a, const void* b) {     return (*(int*)a - *(int*)b); }   /* A helper function that copies contents of arr[]    to Binary Tree. This function basically does Inorder    traversal of Binary Tree and one by one copy arr[]    elements to Binary Tree nodes */ void arrayToBST(int* arr, struct node* root, int* index_ptr) {     // Base Case     if (root == NULL)         return;       /* first update the left subtree */     arrayToBST(arr, root->left, index_ptr);       /* Now update root's data and increment index */     root->data = arr[*index_ptr];     (*index_ptr)++;       /* finally update the right subtree */     arrayToBST(arr, root->right, index_ptr); }   // This function converts a given Binary Tree to BST void binaryTreeToBST(struct node* root) {     // base case: tree is empty     if (root == NULL)         return;       /* Count the number of nodes in Binary Tree so that     we know the size of temporary array to be created */     int n = countNodes(root);       // Create a temp array arr[] and store inorder     // traversal of tree in arr[]     int* arr = new int[n];     int i = 0;     storeInorder(root, arr, &i);       // Sort the array using library function for quick sort     qsort(arr, n, sizeof(arr[0]), compare);       // Copy array elements back to Binary Tree     i = 0;     arrayToBST(arr, root, &i);       // delete dynamically allocated memory to     // avoid memory leak     delete[] arr; }   /* Utility function to create a new Binary Tree node */ struct node* newNode(int data) {     struct node* temp = new struct node;     temp->data = data;     temp->left = NULL;     temp->right = NULL;     return temp; }   /* Utility function to print inorder    traversal of Binary Tree */ void printInorder(struct node* node) {     if (node == NULL)         return;       /* first recur on left child */     printInorder(node->left);       /* then print the data of node */     cout <<" "<< node->data;       /* now recur on right child */     printInorder(node->right); }   /* Driver function to test above functions */ int main() {     struct node* root = NULL;       /* Constructing tree given in the above figure         10         / \         30 15     /     \     20     5 */     root = newNode(10);     root->left = newNode(30);     root->right = newNode(15);     root->left->left = newNode(20);     root->right->right = newNode(5);       // convert Binary Tree to BST     binaryTreeToBST(root);       cout <<"Following is Inorder Traversal of the converted BST:" << endl ;     printInorder(root);       return 0; }    // This code is contributed by shivanisinghss2110

## C

 /* A program to convert Binary Tree to Binary Search Tree */ #include #include   /* A binary tree node structure */ struct node {     int data;     struct node* left;     struct node* right; };   /* A helper function that stores inorder traversal of a tree rooted with node */ void storeInorder(struct node* node, int inorder[], int* index_ptr) {     // Base Case     if (node == NULL)         return;       /* first store the left subtree */     storeInorder(node->left, inorder, index_ptr);       /* Copy the root's data */     inorder[*index_ptr] = node->data;     (*index_ptr)++; // increase index for next entry       /* finally store the right subtree */     storeInorder(node->right, inorder, index_ptr); }   /* A helper function to count nodes in a Binary Tree */ int countNodes(struct node* root) {     if (root == NULL)         return 0;     return countNodes(root->left) + countNodes(root->right) + 1; }   // Following function is needed for library function qsort() int compare(const void* a, const void* b) {     return (*(int*)a - *(int*)b); }   /* A helper function that copies contents of arr[] to Binary Tree. This function basically does Inorder traversal of Binary Tree and one by one copy arr[] elements to Binary Tree nodes */ void arrayToBST(int* arr, struct node* root, int* index_ptr) {     // Base Case     if (root == NULL)         return;       /* first update the left subtree */     arrayToBST(arr, root->left, index_ptr);       /* Now update root's data and increment index */     root->data = arr[*index_ptr];     (*index_ptr)++;       /* finally update the right subtree */     arrayToBST(arr, root->right, index_ptr); }   // This function converts a given Binary Tree to BST void binaryTreeToBST(struct node* root) {     // base case: tree is empty     if (root == NULL)         return;       /* Count the number of nodes in Binary Tree so that     we know the size of temporary array to be created */     int n = countNodes(root);       // Create a temp array arr[] and store inorder traversal of tree in arr[]     int* arr = new int[n];     int i = 0;     storeInorder(root, arr, &i);       // Sort the array using library function for quick sort     qsort(arr, n, sizeof(arr[0]), compare);       // Copy array elements back to Binary Tree     i = 0;     arrayToBST(arr, root, &i);       // delete dynamically allocated memory to avoid memory leak     delete[] arr; }   /* Utility function to create a new Binary Tree node */ struct node* newNode(int data) {     struct node* temp = new struct node;     temp->data = data;     temp->left = NULL;     temp->right = NULL;     return temp; }   /* Utility function to print inorder traversal of Binary Tree */ void printInorder(struct node* node) {     if (node == NULL)         return;       /* first recur on left child */     printInorder(node->left);       /* then print the data of node */     printf("%d ", node->data);       /* now recur on right child */     printInorder(node->right); }   /* Driver function to test above functions */ int main() {     struct node* root = NULL;       /* Constructing tree given in the above figure         10         / \         30 15     /     \     20     5 */     root = newNode(10);     root->left = newNode(30);     root->right = newNode(15);     root->left->left = newNode(20);     root->right->right = newNode(5);       // convert Binary Tree to BST     binaryTreeToBST(root);       printf("Following is Inorder Traversal of the converted BST: \n");     printInorder(root);       return 0; }

## Java

 /* A program to convert Binary Tree to Binary Search Tree */ import java.util.*;   public class GFG{           /* A binary tree node structure */     static class Node {         int data;         Node left;         Node right;     };       // index pointer to pointer to the array index     static int index;         /* A helper function that stores inorder traversal of a tree rooted     with node */     static void storeInorder(Node node, int inorder[])     {         // Base Case         if (node == null)             return;           /* first store the left subtree */         storeInorder(node.left, inorder);           /* Copy the root's data */         inorder[index] = node.data;         index++; // increase index for next entry           /* finally store the right subtree */         storeInorder(node.right, inorder);     }       /* A helper function to count nodes in a Binary Tree */     static int countNodes(Node root)     {         if (root == null)             return 0;         return countNodes(root.left) + countNodes(root.right) + 1;     }       /* A helper function that copies contents of arr[] to Binary Tree.     This function basically does Inorder traversal of Binary Tree and     one by one copy arr[] elements to Binary Tree nodes */     static void arrayToBST(int[] arr, Node root)     {         // Base Case         if (root == null)             return;           /* first update the left subtree */         arrayToBST(arr, root.left);           /* Now update root's data and increment index */         root.data = arr[index];         index++;           /* finally update the right subtree */         arrayToBST(arr, root.right);     }       // This function converts a given Binary Tree to BST     static void binaryTreeToBST(Node root)     {         // base case: tree is empty         if (root == null)             return;           /* Count the number of nodes in Binary Tree so that         we know the size of temporary array to be created */         int n = countNodes(root);           // Create a temp array arr[] and store inorder traversal of tree in arr[]         int arr[] = new int[n];           storeInorder(root, arr);           // Sort the array using library function for quick sort         Arrays.sort(arr);                             // Copy array elements back to Binary Tree         index = 0;         arrayToBST(arr, root);     }       /* Utility function to create a new Binary Tree node */     static Node newNode(int data)     {         Node temp = new Node();         temp.data = data;         temp.left = null;         temp.right = null;         return temp;     }       /* Utility function to print inorder traversal of Binary Tree */     static void printInorder(Node node)     {         if (node == null)             return;           /* first recur on left child */         printInorder(node.left);           /* then print the data of node */         System.out.print(node.data + " ");           /* now recur on right child */         printInorder(node.right);     }       /* Driver function to test above functions */     public static void main(String args[])     {         Node root = null;           /* Constructing tree given in the above figure             10             / \             30 15         /     \         20     5 */         root = newNode(10);         root.left = newNode(30);         root.right = newNode(15);         root.left.left = newNode(20);         root.right.right = newNode(5);           // convert Binary Tree to BST         binaryTreeToBST(root);           System.out.println("Following is Inorder Traversal of the converted BST: ");         printInorder(root);       } }   // This code is contributed by adityapande88.

## Python3

 # Program to convert binary tree to BST   # A binary tree node class Node:           # Constructor to create a new node     def __init__(self, data):         self.data = data         self.left = None         self.right = None   # Helper function to store the inorder traversal of a tree def storeInorder(root, inorder):           # Base Case     if root is None:         return           # First store the left subtree     storeInorder(root.left, inorder)           # Copy the root's data     inorder.append(root.data)       # Finally store the right subtree     storeInorder(root.right, inorder)   # A helper function to count nodes in a binary tree def countNodes(root):     if root is None:         return 0       return countNodes(root.left) + countNodes(root.right) + 1   # Helper function that copies contents of sorted array # to Binary tree def arrayToBST(arr, root):       # Base Case     if root is None:         return           # First update the left subtree     arrayToBST(arr, root.left)       # now update root's data delete the value from array     root.data = arr[0]     arr.pop(0)       # Finally update the right subtree     arrayToBST(arr, root.right)   # This function converts a given binary tree to BST def binaryTreeToBST(root):           # Base Case: Tree is empty     if root is None:         return           # Count the number of nodes in Binary Tree so that     # we know the size of temporary array to be created     n = countNodes(root)       # Create the temp array and store the inorder traversal     # of tree     arr = []     storeInorder(root, arr)           # Sort the array     arr.sort()       # copy array elements back to binary tree     arrayToBST(arr, root)   # Print the inorder traversal of the tree def printInorder(root):     if root is None:         return     printInorder(root.left)     print (root.data,end=" ")     printInorder(root.right)   # Driver program to test above function root = Node(10) root.left = Node(30) root.right = Node(15) root.left.left = Node(20) root.right.right = Node(5)   # Convert binary tree to BST binaryTreeToBST(root)   print ("Following is the inorder traversal of the converted BST") printInorder(root)   # This code is contributed by Nikhil Kumar Singh(nickzuck_007)

## C#

 using System;   public class Node {     public int data;     public Node left;     public Node right; }   public class GFG{           // index pointer to pointer to the array index     static int index;           /* A helper function that stores inorder traversal of a tree rooted     with node */     static void storeInorder(Node node, int[] inorder)     {         // Base Case         if (node == null)             return;            /* first store the left subtree */         storeInorder(node.left, inorder);            /* Copy the root's data */         inorder[index] = node.data;         index++; // increase index for next entry            /* finally store the right subtree */         storeInorder(node.right, inorder);     }        /* A helper function to count nodes in a Binary Tree */     static int countNodes(Node root)     {         if (root == null)             return 0;         return countNodes(root.left) + countNodes(root.right) + 1;     }        /* A helper function that copies contents of arr[] to Binary Tree.     This function basically does Inorder traversal of Binary Tree and     one by one copy arr[] elements to Binary Tree nodes */     static void arrayToBST(int[] arr, Node root)     {         // Base Case         if (root == null)             return;            /* first update the left subtree */         arrayToBST(arr, root.left);            /* Now update root's data and increment index */         root.data = arr[index];         index++;            /* finally update the right subtree */         arrayToBST(arr, root.right);     }        // This function converts a given Binary Tree to BST     static void binaryTreeToBST(Node root)     {         // base case: tree is empty         if (root == null)             return;            /* Count the number of nodes in Binary Tree so that         we know the size of temporary array to be created */         int n = countNodes(root);            // Create a temp array arr[] and store inorder traversal of tree in arr[]         int[] arr = new int[n];            storeInorder(root, arr);            // Sort the array using library function for quick sort         Array.Sort(arr);                               // Copy array elements back to Binary Tree         index = 0;         arrayToBST(arr, root);     }        /* Utility function to create a new Binary Tree node */     static Node newNode(int data)     {         Node temp = new Node();         temp.data = data;         temp.left = null;         temp.right = null;         return temp;     }        /* Utility function to print inorder traversal of Binary Tree */     static void printInorder(Node node)     {         if (node == null)             return;            /* first recur on left child */         printInorder(node.left);            /* then print the data of node */         Console.Write(node.data + " ");            /* now recur on right child */         printInorder(node.right);     }        /* Driver function to test above functions */           static public void Main (){                   Node root = null;            /* Constructing tree given in the above figure             10             / \             30 15         /     \         20     5 */         root = newNode(10);         root.left = newNode(30);         root.right = newNode(15);         root.left.left = newNode(20);         root.right.right = newNode(5);            // convert Binary Tree to BST         binaryTreeToBST(root);            Console.WriteLine("Following is Inorder Traversal of the converted BST: ");         printInorder(root);               } }   // This code is contributed by avanitrachhadiya2155

## Javascript



Output

Following is the inorder traversal of the converted BST
5 10 15 20 30

Complexity Analysis:

• Time Complexity: O(nlogn). This is the complexity of the sorting algorithm which we are using after first in-order traversal, rest of the operations take place in linear time.
• Auxiliary Space: O(n). Use of data structure ‘array’ to store in-order traversal.

Another approach :- Using Inorder Traversal and Vector Sorting

In this approach, we will first perform an inorder traversal of the binary tree and store the nodes in a vector. After that, we will sort the vector in ascending order, and then use the sorted vector to construct a binary search tree.

Define a struct for a binary tree node with a value, left and right pointers.
Implement an inorder traversal of the binary tree to store the nodes in a vector. In this implementation, a vector of TreeNode* is used to store the nodes in inorder traversal order. The function takes in a TreeNode* root and a reference to the vector of TreeNode* nodes.
Implement a function to construct a binary search tree from the sorted vector of TreeNode* nodes. The function takes in the vector of TreeNode* nodes, start index, and end index of the vector. It recursively constructs the binary search tree by taking the middle element of the vector as the root, and then constructing the left and right subtrees recursively.
Implement a function to convert a binary tree to a binary search tree. This function takes in the root of the binary tree and returns the root of the binary search tree. It first calls the inorder traversal function to store the nodes of the binary tree in a vector, then sorts the vector of TreeNode* based on the values of the nodes using a lambda function, and finally calls the function to construct the binary search tree.
Implement a function to print the inorder traversal of a binary tree. This function takes in the root of the binary tree and prints its inorder traversal.
In the driver code, create a binary tree with some nodes, print its inorder traversal, convert it to a binary search tree, print the inorder traversal of the binary search tree, and free the dynamically allocated memory for the nodes.

Here’s the C++ code:

## C++

 #include #include #include using namespace std;   // Definition for a binary tree node. struct TreeNode {     int val;     TreeNode* left;     TreeNode* right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {} };   // Inorder traversal to store the nodes in a vector void inorder(TreeNode* root, vector& nodes) {     if (root == NULL) {         return;     }     inorder(root->left, nodes);     nodes.push_back(root);     inorder(root->right, nodes); }   // Function to construct a binary search tree from a sorted vector TreeNode* constructBST(vector& nodes, int start, int end) {     if (start > end) {         return NULL;     }     int mid = (start + end) / 2;     TreeNode* root = nodes[mid];     root->left = constructBST(nodes, start, mid - 1);     root->right = constructBST(nodes, mid + 1, end);     return root; }   // Function to convert a binary tree to a binary search tree TreeNode* convertToBST(TreeNode* root) {     vector nodes;     inorder(root, nodes);     sort(nodes.begin(), nodes.end(), [](const TreeNode* a, const TreeNode* b) {         return a->val < b->val;     });     return constructBST(nodes, 0, nodes.size() - 1); }   // Function to print the inorder traversal of a binary tree void printInorder(TreeNode* root) {     if (root == NULL) {         return;     }     printInorder(root->left);     cout << root->val << " ";     printInorder(root->right); }   // Driver code int main() {     // Example binary tree     TreeNode* root = new TreeNode(10);     root->left = new TreeNode(30);     root->right = new TreeNode(15);     root->left->left = new TreeNode(20);     root->left->right = new TreeNode(5);             // Convert binary tree to binary search tree     TreeNode* bst = convertToBST(root);     cout <<"Following is Inorder Traversal of the converted BST:" << endl ;     printInorder(bst);     cout << endl;       return 0; }

## Java

 import java.util.ArrayList; import java.util.Collections;   class TreeNode {   int val;   TreeNode left;   TreeNode right;     TreeNode(int x) {     val = x;   } }   public class Main {   public static void main(String[] args) {     // Example binary tree     TreeNode root = new TreeNode(10);     root.left = new TreeNode(30);     root.right = new TreeNode(15);     root.left.left = new TreeNode(20);     root.left.right = new TreeNode(5);       // Convert binary tree to binary search tree     TreeNode bst = convertToBST(root);       System.out.println("Following is Inorder Traversal of the converted BST:");     printInorder(bst);   }     // Inorder traversal to store the nodes in an ArrayList   public static void inorder(TreeNode root, ArrayList nodes) {     if (root == null) {       return;     }     inorder(root.left, nodes);     nodes.add(root);     inorder(root.right, nodes);   }     // Function to construct a binary search tree from a sorted ArrayList   public static TreeNode constructBST(ArrayList nodes, int start, int end) {     if (start > end) {       return null;     }     int mid = (start + end) / 2;     TreeNode root = nodes.get(mid);     root.left = constructBST(nodes, start, mid - 1);     root.right = constructBST(nodes, mid + 1, end);     return root;   }     // Function to convert a binary tree to a binary search tree   public static TreeNode convertToBST(TreeNode root) {     ArrayList nodes = new ArrayList<>();     inorder(root, nodes);     Collections.sort(nodes, (a, b) -> a.val - b.val);     return constructBST(nodes, 0, nodes.size() - 1);   }     // Function to print the inorder traversal of a binary tree   public static void printInorder(TreeNode root) {     if (root == null) {       return;     }     printInorder(root.left);     System.out.print(root.val + " ");     printInorder(root.right);   } }

Time Complexity: O(nlogn).
Auxiliary Space: O(n).

We will be covering another method for this problem which converts the tree using O(height of the tree) extra space.

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