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 C implementation of the above approach. The main function to convert is highlighted in the following code.
C
/* A program to convert Binary Tree to Binary Search Tree */#include<stdio.h> #include<stdlib.h> /* 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 functon 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 meory 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:
"); printInorder (root); return 0; } |
Python
# 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 inroder 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 funtion 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 traveral # 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, 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) |
Output:
Following is Inorder Traversal of the converted BST: 5 10 15 20 30
We will be covering another method for this problem which converts the tree using O(height of tree) extra space.
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