Question

(Python) 

Write a python function that takes a binary search T and a range [a, b] with a ≤ b, where a,b are two integers, and has to decide whether there are two different keys k1,k2 in T whose sum belongs to the given range [a, b] or not. For instance, if the input to the function is some BST T and the range [−1, 4], the function has to decide whether there are two distinct keys k1 , k2 in T whose sum is in the range [−1, 4], where −1 and 4 are included in the range. As an another example, if the input to the function is some BST T and the range [0, 0], the function has to decide whether or not there are two distinct keys k1 , k2 in T whose sum is 0. The function has to return TRUE if there are two distinct keys whose sum belongs to the given range. And, the function has to return FALSE otherwise. The function has to be started as the following:

def BSTRangeSumSearch(T: BinaryTree, a: int, b: int) -> bool:

Answer 1

Write a python function that takes a binary search T and a range [a, b] with a ≤ b, where a,b are two integers, and has to decide whether there are two different keys k1,k2 in T whose sum belongs to the given range [a, b] or not. For instance, if the input to the function is some BST T and the range [−1, 4], the function has to decide whether there are two distinct keys k1 , k2 in T whose sum is in the range [−1, 4], where −1 and 4 are included in the range. As an another example, if the input to the function is some BST T and the range [0, 0], the function has to decide whether or not there are two distinct keys k1 , k2 in T whose sum is 0. The function has to return TRUE if there are two distinct keys whose sum belongs to the given range. And, the function has to return FALSE otherwise. The function has to be started as the following:

def BSTRangeSumSearch(T: BinaryTree, a: int, b: int) -> bool:

# 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

# The function prints all the keys in the gicven range
# [k1..k2]. Assumes that k1 < k2
def Print(root, k1, k2):
  
   # Base Case
   if root is None:
       return

   # Since the desired o/p is sorted, recurse for left
   # subtree first. If root.data is greater than k1, then
   # only we can get o/p keys in left subtree
   if k1 < root.data :
       Print(root.left, k1, k2)

   # If root's data lies in range, then prints root's data
   if k1 <= root.data and k2 >= root.data:
       print root.data,

   # If root.data is smaller than k2, then only we can get
   # o/p keys in right subtree
   if k2 > root.data:
       Print(root.right, k1, k2)

# Driver function to test above function
k1 = 10 ; k2 = 25 ;
root = Node(20)
root.left = Node(8)
root.right = Node(22)
root.left.left = Node(4)
root.left.right = Node(12)

Print(root, k1, k2)
Output:

12 20 22

Time Complexity: O(n) where n is the total number of keys in tree.