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Split a BST into two balanced BSTs based on a value K
  • Difficulty Level : Hard
  • Last Updated : 02 Nov, 2020

Given a Binary Search tree and an integer K, we have to split the tree into two Balanced Binary Search Tree, where BST-1 consists of all the nodes which are less than K and BST-2 consists of all the nodes which are greater than or equal to K.
Note: Arrangement of the nodes may be anything but both BST should be Balanced.
Examples: 

Input:
         40                            
        /   \    
      20     50     
     /  \      \               
    10   35     60
        /      /   
      25      55 
K = 35
Output:
First BST: 10 20 25
Second BST: 35 40 50 55 60
Explanation:
After splitting above BST
about given value K = 35
First Balanced Binary Search Tree is 
         20                            
        /   \    
      10     25 
Second Balanced Binary Search Tree is
         50                            
        /   \    
      35     55     
        \      \               
         40     60
OR
         40                            
        /   \    
      35     55     
            /   \               
           50    60

Input:
         100                            
        /   \    
      20     500     
     /  \                     
    10   30     
           \      
            40 
K = 50
Output:
First BST: 10 20 30 40
Second BST: 100 500
Explanation:
After splitting above BST 
about given value K = 50
First Balanced Binary Search Tree is 
         20                            
        /   \    
      10     30 
                \
                 40
Second Balanced Binary Search Tree is
         100                            
            \    
             500     

Approach:  

  1. First store the inorder traversal of given BST in an array
  2. Then, split this array about given value K
  3. Now construct first balanced BST by first splitting part and second BST by second splitting part, using the approach used in this article.
     

Below is the implementation of the above approach:
 

C++




// C++ program to split a BST into
// two balanced BSTs based on a value K
 
#include <iostream>
using namespace std;
 
// Structure of each node of BST
struct node {
    int key;
    struct node *left, *right;
};
 
// A utility function to
// create a new BST node
node* newNode(int item)
{
    node* temp = new node();
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}
 
// A utility function to insert a new
// node with given key in BST
struct node* insert(struct node* node,
                    int key)
{
    // If the tree is empty, return a new node
    if (node == NULL)
        return newNode(key);
 
    // Otherwise, recur down the tree
    if (key < node->key)
        node->left = insert(node->left,
                            key);
    else if (key > node->key)
        node->right = insert(node->right,
                             key);
 
    // return the (unchanged) node pointer
    return node;
}
 
// Function to return the size
// of the tree
int sizeOfTree(node* root)
{
    if (root == NULL) {
        return 0;
    }
 
    // Calculate left size recursively
    int left = sizeOfTree(root->left);
 
    // Calculate right size recursively
    int right = sizeOfTree(root->right);
 
    // Return total size recursively
    return (left + right + 1);
}
 
// Function to store inorder
// traversal of BST
void storeInorder(node* root,
                  int inOrder[],
                  int& index)
{
    // Base condition
    if (root == NULL) {
        return;
    }
 
    // Left recursive call
    storeInorder(root->left,
                 inOrder, index);
 
    // Store elements in inorder array
    inOrder[index++] = root->key;
 
    // Right recursive call
    storeInorder(root->right,
                 inOrder, index);
}
 
// Function to return the splitting
// index of the array
int getSplittingIndex(int inOrder[],
                      int index, int k)
{
    for (int i = 0; i < index; i++) {
        if (inOrder[i] >= k) {
            return i - 1;
        }
    }
    return index - 1;
}
 
// Function to create the Balanced
// Binary search tree
node* createBST(int inOrder[],
                int start, int end)
{
    // Base Condition
    if (start > end) {
        return NULL;
    }
 
    // Calculate the mid of the array
    int mid = (start + end) / 2;
    node* t = newNode(inOrder[mid]);
 
    // Recursive call for left child
    t->left = createBST(inOrder,
                        start, mid - 1);
 
    // Recursive call for right child
    t->right = createBST(inOrder,
                         mid + 1, end);
 
    // Return newly created Balanced
    // Binary Search Tree
    return t;
}
 
// Function to traverse the tree
// in inorder fashion
void inorderTrav(node* root)
{
    if (root == NULL)
        return;
    inorderTrav(root->left);
    cout << root->key << " ";
    inorderTrav(root->right);
}
 
// Function to split the BST
// into two Balanced BST
void splitBST(node* root, int k)
{
 
    // Print the original BST
    cout << "Original BST : ";
    if (root != NULL) {
        inorderTrav(root);
    }
    else {
        cout << "NULL";
    }
    cout << endl;
 
    // Store the size of BST1
    int numNode = sizeOfTree(root);
 
    // Take auxiliary array for storing
    // The inorder traversal of BST1
    int inOrder[numNode + 1];
    int index = 0;
 
    // Function call for storing
    // inorder traversal of BST1
    storeInorder(root, inOrder, index);
 
    // Function call for getting
    // splitting index
    int splitIndex
        = getSplittingIndex(inOrder,
                            index, k);
 
    node* root1 = NULL;
    node* root2 = NULL;
 
    // Creation of first Balanced
    // Binary Search Tree
    if (splitIndex != -1)
        root1 = createBST(inOrder, 0,
                          splitIndex);
 
    // Creation of Second Balanced
    // Binary Search Tree
    if (splitIndex != (index - 1))
        root2 = createBST(inOrder,
                          splitIndex + 1,
                          index - 1);
 
    // Print two Balanced BSTs
    cout << "First BST : ";
    if (root1 != NULL) {
        inorderTrav(root1);
    }
    else {
        cout << "NULL";
    }
    cout << endl;
 
    cout << "Second BST : ";
    if (root2 != NULL) {
        inorderTrav(root2);
    }
    else {
        cout << "NULL";
    }
}
 
// Driver code
int main()
{
    /*  BST
             5
           /   \     
          3     7    
         / \   / \   
         2  4  6  8 
    */
    struct node* root = NULL;
    root = insert(root, 5);
    insert(root, 3);
    insert(root, 2);
    insert(root, 4);
    insert(root, 7);
    insert(root, 6);
    insert(root, 8);
 
    int k = 5;
 
    // Function to split BST
    splitBST(root, k);
 
    return 0;
}

Python3




# Python 3 program to split a
# BST into two balanced BSTs
# based on a value K
index = 0
 
# Structure of each node of BST
class newNode:
    def __init__(self, item):
       
        # A utility function to
        # create a new BST node
        self.key = item
        self.left = None
        self.right = None
 
# A utility function to insert
# a new node with given key
# in BST
def insert(node, key):
   
    # If the tree is empty,
    # return a new node
    if (node == None):
        return newNode(key)
 
    # Otherwise, recur down
    # the tree
    if (key < node.key):
        node.left = insert(node.left,
                           key)
    elif (key > node.key):
        node.right = insert(node.right,
                            key)
 
    # return the (unchanged)
    # node pointer
    return node
 
# Function to return the
# size of the tree
def sizeOfTree(root):
   
    if (root == None):
        return 0
 
    # Calculate left size
    # recursively
    left = sizeOfTree(root.left)
 
    # Calculate right size
    # recursively
    right = sizeOfTree(root.right)
 
    # Return total size
    # recursively
    return (left + right + 1)
 
# Function to store inorder
# traversal of BST
def storeInorder(root, inOrder):
   
    global index
    # Base condition
    if (root == None):
        return
 
    # Left recursive call
    storeInorder(root.left,
                 inOrder)
 
    # Store elements in
    # inorder array
    inOrder[index] = root.key
    index += 1
 
    # Right recursive call
    storeInorder(root.right,
                 inOrder)
 
# Function to return the
# splitting index of the
# array
def getSplittingIndex(inOrder,
                      index, k):
   
    for i in range(index):
        if (inOrder[i] >= k):
            return i - 1
    return index - 1
 
# Function to create the
# Balanced Binary search
# tree
def createBST(inOrder,
              start, end):
   
    # Base Condition
    if (start > end):
        return None
 
    # Calculate the mid of
    # the array
    mid = (start + end) // 2
    t = newNode(inOrder[mid])
 
    # Recursive call for
    # left child
    t.left = createBST(inOrder,
                       start,
                       mid - 1)
 
    # Recursive call for
    # right child
    t.right = createBST(inOrder,
                        mid + 1, end)
 
    # Return newly created
    # Balanced Binary Search
    # Tree
    return t
 
# Function to traverse
# the tree in inorder
# fashion
def inorderTrav(root):
   
    if (root == None):
        return
       
    inorderTrav(root.left)
    print(root.key, end = " ")
    inorderTrav(root.right)
 
# Function to split the BST
# into two Balanced BST
def splitBST(root, k):
   
    global index
     
    # Print the original BST
    print("Original BST : ")
    if (root != None):
        inorderTrav(root)
        print("\n", end = "")
    else:
        print("NULL")
 
    # Store the size of BST1
    numNode = sizeOfTree(root)
 
    # Take auxiliary array for
    # storing The inorder traversal
    # of BST1
    inOrder = [0 for i in range(numNode + 1)]
    index = 0
 
    # Function call for storing
    # inorder traversal of BST1
    storeInorder(root, inOrder)
 
    # Function call for getting
    # splitting index
    splitIndex = getSplittingIndex(inOrder,
                                   index, k)
 
    root1 = None
    root2 = None
 
    # Creation of first Balanced
    # Binary Search Tree
    if (splitIndex != -1):
        root1 = createBST(inOrder,
                          0, splitIndex)
 
    # Creation of Second Balanced
    # Binary Search Tree
    if (splitIndex != (index - 1)):
        root2 = createBST(inOrder,
                          splitIndex + 1,
                          index - 1)
 
    # Print two Balanced BSTs
    print("First BST : ")
    if (root1 != None):
        inorderTrav(root1)
        print("\n", end = "")
    else:
        print("NULL")
 
    print("Second BST : ")
    if (root2 != None):
        inorderTrav(root2)
        print("\n", end = "")
    else:
        print("NULL")
 
# Driver code
if __name__ == '__main__':
   
    '''/*  BST
             5
           /   /     
          3     7    
         / /   / /   
         2  4  6  8 
    */'''
    root = None
    root = insert(root, 5)
    insert(root, 3)
    insert(root, 2)
    insert(root, 4)
    insert(root, 7)
    insert(root, 6)
    insert(root, 8)
 
    k = 5
 
    # Function to split BST
    splitBST(root, k)
 
# This code is contributed by Chitranayal
Output: 
Original BST : 2 3 4 5 6 7 8 
First BST : 2 3 4 
Second BST : 5 6 7 8


 




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