Check if the Binary Tree contains a balanced BST of size K

Given a Binary Tree and a positive integer K. The task is to check whether the Balanced BST of size K exists in a given Binary Tree or not. If it exists then print “Yes” else print “No”.

Examples: 

Input: K = 4,
Below is the given Tree:
         15
       /    \
      10     26
     /  \     / \
    5   12  25  40
           /   /  \
          20  35   50
                 \
                  60
Output: Yes
Explanation: 
Subtree of the given tree with
size k is given below:
        40
       /  \
      35   50
             \
             60

Input: K = 4,
Below is the given Tree:
            18
            /  
           9    
          / \
         7   10
Output: No
Explanation:
There is no subtree of size K
which forms a balanced BT.

Approach: The idea is to use the Post Order Traversal. The following are the steps for solving the problem: 

  1. Perform a Post Order Traversal on the given tree and check BST condition for each node where the largest value in the left sub-tree should be smaller than the current value and the smaller value in the right subtree should be greater than the current value.
  2. Then check if the BST is balanced or not that is the absolute difference between left and right sub-tree should be either 0 or 1.
  3. Then pass values return from the sub-trees to the parent.
  4. Perform the above steps for all nodes and take the boolean variable ans which is initially marked false which is used to check whether a balanced BST is present or not.
  5. If a Balanced BST of size K is found then print “Yes” else print “No”.

Below is the implementation of the above approach:

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// A tree node
struct node {
    int data;
    node* left;
    node* right;
};
 
// Structure of temporary variable
struct minMax {
    bool isBST;
    bool balanced;
    int size;
    int hight;
    int min;
    int max;
};
 
// Function to create the node
node* createNode(int value)
{
    node* temp = new node();
    temp->left = NULL;
    temp->right = NULL;
    temp->data = value;
}
 
// Utility function to find Balanced
// BST of size k
minMax findBalancedBstUtil(node* root,
                           int k, bool& ans)
{
    // Base condition
    if (root == NULL)
        return { true, true, 0, 0,
                 INT_MAX, INT_MIN };
 
    // Temporary variable
    minMax temp;
 
    // Recursive call for left sub-tree
    minMax lsTree
        = findBalancedBstUtil(root->left,
                              k, ans);
 
    if (ans == true)
        return temp;
 
    // Recursive call for right sub-tree
    minMax rsTree
        = findBalancedBstUtil(root->right,
                              k, ans);
 
    if (ans == true)
        return temp;
 
    // Check those conditions which
    // violated the rules of BST
    if (!lsTree.isBST || !rsTree.isBST
        || lsTree.max > root->data
        || rsTree.min < root->data) {
        temp.isBST = false;
        return temp;
    }
 
    // Check whether the Bst is
    // height balanced or not
    if (abs(lsTree.hight
            - rsTree.hight)
            == 1
        || abs(lsTree.hight
               - rsTree.hight)
               == 0)
 
        temp.balanced = true;
 
    else
        temp.balanced = false;
 
    // Make the variable true
    // as sub-tree is BST
    temp.isBST = true;
 
    // Store the size
    temp.size = 1 + lsTree.size
                + rsTree.size;
 
    // Store the height
    temp.hight = max(lsTree.hight,
                     rsTree.hight)
                 + 1;
 
    // Store the minimum of BST
    temp.min = root->left != NULL
                   ? lsTree.min
                   : root->data;
 
    // Store the maximum of BST
    temp.max = root->right != NULL
                   ? rsTree.max
                   : root->data;
 
    // Condition to check whether the
    // size of Balnced BST is K or not
    if (temp.balanced == true
        && temp.size == k) {
        ans = true;
    }
 
    // Return the temporary variable
    // with updated data
    return temp;
}
 
// Function to find the Balanced
// BST of size k
string findBalancedBst(node* root,
                       int k)
{
    bool ans = false;
 
    // Utility function call
    findBalancedBstUtil(root, k, ans);
    return ans == true ? "Yes" : "No";
}
 
// Driver Code
int main()
{
    // Given Binary Tree
    node* root = createNode(15);
    root->left = createNode(10);
    root->right = createNode(26);
    root->left->left = createNode(5);
    root->left->right = createNode(12);
    root->right->left = createNode(25);
    root->right->left->left = createNode(20);
    root->right->right = createNode(40);
    root->right->right->left = createNode(35);
    root->right->right->right = createNode(50);
    root->right->right->right->right = createNode(60);
 
    int k = 4;
 
    // Function Call
    cout << findBalancedBst(root, k);
    return 0;
}
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// Java program for the above approach
import java.util.*;
 
class GFG{
     
static boolean ans;
 
// A tree node
static class node
{
    int data;
    node left;
    node right;
};
 
// Structure of temporary variable
static class minMax
{
    boolean isBST;
    boolean balanced;
    int size;
    int hight;
    int min;
    int max;
     
    public minMax(boolean isBST, boolean balanced,
                  int size, int hight, int min,
                  int max)
    {
        super();
        this.isBST = isBST;
        this.balanced = balanced;
        this.size = size;
        this.hight = hight;
        this.min = min;
        this.max = max;
    }
    public minMax()
    {
        // TODO Auto-generated constructor stub
    }
};
 
// Function to create the node
static node createNode(int value)
{
    node temp = new node();
    temp.left = null;
    temp.right = null;
    temp.data = value;
    return temp;
}
 
// Utility function to find Balanced
// BST of size k
static minMax findBalancedBstUtil(node root,
                                  int k)
{
     
    // Base condition
    if (root == null)
        return new minMax(true, true, 0, 0,
                          Integer.MAX_VALUE,
                          Integer.MIN_VALUE );
 
    // Temporary variable
    minMax temp = new minMax();
 
    // Recursive call for left sub-tree
    minMax lsTree = findBalancedBstUtil(root.left,
                                        k);
 
    if (ans == true)
        return temp;
 
    // Recursive call for right sub-tree
    minMax rsTree = findBalancedBstUtil(root.right,
                                        k);
 
    if (ans == true)
        return temp;
 
    // Check those conditions which
    // violated the rules of BST
    if (!lsTree.isBST || !rsTree.isBST ||
         lsTree.max > root.data ||
         rsTree.min < root.data)
    {
        temp.isBST = false;
        return temp;
    }
 
    // Check whether the Bst is
    // height balanced or not
    if (Math.abs(lsTree.hight -
                 rsTree.hight) == 1 ||
        Math.abs(lsTree.hight -
                 rsTree.hight) == 0)
        temp.balanced = true;
         
    else
        temp.balanced = false;
 
    // Make the variable true
    // as sub-tree is BST
    temp.isBST = true;
 
    // Store the size
    temp.size = 1 + lsTree.size +
                    rsTree.size;
 
    // Store the height
    temp.hight = Math.max(lsTree.hight,
                          rsTree.hight) + 1;
 
    // Store the minimum of BST
    temp.min = root.left != null ?
               lsTree.min : root.data;
 
    // Store the maximum of BST
    temp.max = root.right != null ?
               rsTree.max : root.data;
 
    // Condition to check whether the
    // size of Balnced BST is K or not
    if (temp.balanced == true &&
            temp.size == k)
    {
        ans = true;
    }
 
    // Return the temporary variable
    // with updated data
    return temp;
}
 
// Function to find the Balanced
// BST of size k
static String findBalancedBst(node root,
                              int k)
{
    ans = false;
 
    // Utility function call
    findBalancedBstUtil(root, k);
    return ans == true ? "Yes" : "No";
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Binary Tree
    node root = createNode(15);
    root.left = createNode(10);
    root.right = createNode(26);
    root.left.left = createNode(5);
    root.left.right = createNode(12);
    root.right.left = createNode(25);
    root.right.left.left = createNode(20);
    root.right.right = createNode(40);
    root.right.right.left = createNode(35);
    root.right.right.right = createNode(50);
    root.right.right.right.right = createNode(60);
 
    int k = 4;
 
    // Function call
    System.out.print(findBalancedBst(root, k));
}
}
 
// This code is contributed by Amit Katiyar
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# Python3 program for the above approach
import sys
 
ans = False
 
# A tree node
class createNode:
     
    def __init__(self, data):
         
        self.data = data
        self.left = None
        self.right = None
 
# Structure of temporary variable
class newMinMax:
     
    def __init__(self, isBST, balanced, size,
                 height, mn, mx):
                      
        self.isBST = isBST
        self.balanced = balanced
        self.size = size
        self.height = height
        self.mn = mn
        self.mx = mx
 
# Utility function to find Balanced
# BST of size k
def findBalancedBstUtil(root, k):
     
    global ans
     
    # Base condition
    if (root == None):
        return newMinMax(True, True, 0, 0,
                         sys.maxsize,
                        -sys.maxsize - 1)
 
    # Temporary variable
    temp = newMinMax(True, True, 0, 0,
                     sys.maxsize,
                    -sys.maxsize - 1)
 
    # Recursive call for left sub-tree
    lsTree = findBalancedBstUtil(root.left, k)
 
    if (ans == True):
        return temp
 
    # Recursive call for right sub-tree
    rsTree = findBalancedBstUtil(root.right, k)
 
    if (ans == True):
        return temp
 
    # Check those conditions which
    # violated the rules of BST
    if (lsTree.isBST == False or
        rsTree.isBST == False or
        lsTree.mx > root.data or
        rsTree.mn < root.data):
        temp.isBST = False
        return temp
 
    # Check whether the Bst is
    # height balanced or not
    if (abs(lsTree.height - rsTree.height) == 1 or
        abs(lsTree.height - rsTree.height) == 0):
        temp.balanced = True
    else:
        temp.balanced = False
 
    # Make the variable true
    # as sub-tree is BST
    temp.isBST = True
 
    # Store the size
    temp.size = 1 + lsTree.size + rsTree.size
 
    # Store the height
    temp.height  = max(lsTree.height ,
                       rsTree.height) + 1
 
    # Store the minimum of BST
    if root.left != None:
        temp.mn = lsTree.mn
    else:
        temp.mn = root.data
 
    # Store the maximum of BST
    if root.right != None:
        temp.mx = rsTree.mx
    else:
        temp.mx = root.data
 
    # Condition to check whether the
    # size of Balnced BST is K or not
    if (temp.balanced == True and
        temp.size == k):
        ans = True
 
    # Return the temporary variable
    # with updated data
    return temp
 
# Function to find the Balanced
# BST of size k
def findBalancedBst(root, k):
     
    global ans
     
    # Utility function call
    findBalancedBstUtil(root, k)
    if ans == True:
        return "Yes"
    else:
        return "No"
 
# Driver Code
if __name__ == '__main__':
     
    # Given Binary Tree
    root = createNode(15)
    root.left = createNode(10)
    root.right = createNode(26)
    root.left.left = createNode(5)
    root.left.right = createNode(12)
    root.right.left = createNode(25)
    root.right.left.left = createNode(20)
    root.right.right = createNode(40)
    root.right.right.left = createNode(35)
    root.right.right.right = createNode(50)
    root.right.right.right.right = createNode(60)
 
    k = 4
 
    # Function Call
    print(findBalancedBst(root, k))
 
# This code is contributed by ipg2016107
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// C# program for the
// above approach
using System;
class GFG{
     
static bool ans;
 
// A tree node
public class node
{
  public int data;
  public node left;
  public node right;
};
 
// Structure of temporary
// variable
public class minMax
{
  public bool isBST;
  public bool balanced;
  public int size;
  public int hight;
  public int min;
  public int max;   
  public minMax(bool isBST, bool balanced,
                int size, int hight, int min,
                int max)
  {
    this.isBST = isBST;
    this.balanced = balanced;
    this.size = size;
    this.hight = hight;
    this.min = min;
    this.max = max;
  }
   
  public minMax()
  {
    // TODO Auto-generated constructor stub
  }
};
 
// Function to create the node
static node createNode(int value)
{
  node temp = new node();
  temp.left = null;
  temp.right = null;
  temp.data = value;
  return temp;
}
 
// Utility function to find Balanced
// BST of size k
static minMax findBalancedBstUtil(node root,
                                  int k)
{
  // Base condition
  if (root == null)
    return new minMax(true, true, 0, 0,
                      int.MaxValue,
                      int.MinValue);
 
  // Temporary variable
  minMax temp = new minMax();
 
  // Recursive call for left sub-tree
  minMax lsTree =
         findBalancedBstUtil(root.left, k);
 
  if (ans == true)
    return temp;
 
  // Recursive call for right sub-tree
  minMax rsTree =
         findBalancedBstUtil(root.right, k);
 
  if (ans == true)
    return temp;
 
  // Check those conditions which
  // violated the rules of BST
  if (!lsTree.isBST || !rsTree.isBST ||
      lsTree.max > root.data ||
      rsTree.min < root.data)
  {
    temp.isBST = false;
    return temp;
  }
 
  // Check whether the Bst is
  // height balanced or not
  if (Math.Abs(lsTree.hight -
               rsTree.hight) == 1 ||
      Math.Abs(lsTree.hight -
               rsTree.hight) == 0)
    temp.balanced = true;
 
  else
    temp.balanced = false;
 
  // Make the variable true
  // as sub-tree is BST
  temp.isBST = true;
 
  // Store the size
  temp.size = 1 + lsTree.size +
                  rsTree.size;
 
  // Store the height
  temp.hight = Math.Max(lsTree.hight,
                        rsTree.hight) + 1;
 
  // Store the minimum of BST
  temp.min = root.left != null ?
             lsTree.min : root.data;
 
  // Store the maximum of BST
  temp.max = root.right != null ?
             rsTree.max : root.data;
 
  // Condition to check whether the
  // size of Balnced BST is K or not
  if (temp.balanced == true &&
      temp.size == k)
  {
    ans = true;
  }
 
  // Return the temporary
  // variable with updated data
  return temp;
}
 
// Function to find the Balanced
// BST of size k
static String findBalancedBst(node root,
                              int k)
{
  ans = false;
 
  // Utility function call
  findBalancedBstUtil(root, k);
  return ans == true ? "Yes" : "No";
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given Binary Tree
  node root = createNode(15);
  root.left = createNode(10);
  root.right = createNode(26);
  root.left.left = createNode(5);
  root.left.right = createNode(12);
  root.right.left = createNode(25);
  root.right.left.left = createNode(20);
  root.right.right = createNode(40);
  root.right.right.left = createNode(35);
  root.right.right.right = createNode(50);
  root.right.right.right.right = createNode(60);
 
  int k = 4;
 
  // Function call
  Console.Write(findBalancedBst(root, k));
}
}
 
// This code is contributed by Princi Singh
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Output: 
Yes









 

Time Complexity: O(N) 
Auxiliary Space: O(1)

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