Find if given vertical level of binary tree is sorted or not in Python

Suppose we have binary tree; we have to check whether the given vertical level of the binary tree is sorted or not. When two nodes are overlapping, check they are in sorted sequence in the level they belong.

So, if the input is like l = -1

then the output will be True, as elements in level -1 are 3,7 and they are sorted.

To solve this, we will follow these steps −

• if root is null, then

  • return True

• previous_value := -inf

• current_level := 0

• current_node := a new tree node with value 0

• Define one dequeue named q

• insert (root, 0) at the end of q

• while q is not empty, do

  • current_node := q[0, 0]

  • current_level := q[0, 1]

  • delete element from left of q

  • if current_level is same as level, then

    • if previous_value <= current_node.val, then

      • previous_value := current_node.val

    • otherwise,

      • return False

  • if current_node.left is not-null, then

    • insert (current_node.left, current_level - 1) at the end of q

  • if current_node.right is non-null, then

    • insert (current_node.right, current_level + 1) at the end of q

• return True

Example

Let us see the following implementation to get better understanding −

 Live Demo

from collections import deque
from sys import maxsize
INT_MIN = -maxsize
class TreeNode:
   def __init__(self, key):
      self.val = key
      self.left = None
      self.right = None
def are_elements_sorted(root, level):
   if root is None:
      return True
   previous_value = INT_MIN
   current_level = 0
   current_node = TreeNode(0)
   q = deque()
   q.append((root, 0))
   while q:
      current_node = q[0][0]
      current_level = q[0][1]
      q.popleft()
      if current_level == level:
         if previous_value <= current_node.val:
            previous_value = current_node.val
         else:
            return False
      if current_node.left:
         q.append((current_node.left, current_level - 1))
      if current_node.right:
         q.append((current_node.right, current_level + 1))
   return True
root = TreeNode(2)
root.left = TreeNode(3)
root.right = TreeNode(6)
root.left.left = TreeNode(8)
root.left.right = TreeNode(5)
root.left.right.left = TreeNode(7)
level = -1
print(are_elements_sorted(root, level))

Input

root = TreeNode(2) root.left = TreeNode(3) root.right = TreeNode(6)
root.left.left = TreeNode(8) root.left.right = TreeNode(5)
root.left.right.left = TreeNode(7)

Output

True