<filename>is_balanced.py
"""
Given a Binary Tree (BT), convert it to a Doubly Linked List(DLL) In-Place. The left and right pointers in nodes are to be used as previous 
and next pointers respectively in converted DLL. The order of nodes in DLL must be same as Inorder of the given Binary Tree. The first node 
of Inorder traversal (leftmost node in BT) must be the head node of the DLL.

TreeToList

Example 1:

Input:
      1
    /  \
   3    2
Output:
3 1 2 
2 1 3 
Explanation: DLL would be 3<=>1<=>2
Example 2:

Input:
       10
      /   \
     20   30
   /   \
  40   60
Output:
40 20 60 10 30 
30 10 60 20 40
Explanation:  DLL would be 
40<=>20<=>60<=>10<=>30.
Your Task:
You don't have to take input. Complete the function bToDLL() that takes root node of the tree as a parameter and returns the head of DLL. 
The driver code prints the DLL both ways.

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(H).
Note: H is the height of the tree and this space is used implicitly for recursion stack.
"""
from math import abs, max


class Solution:
    def __init__(self) -> None:
        self.height = 0

    def isBalanced(self, root_node):
        if root_node is None:
            return True

        left = self.isBalanced(root_node.left)
        right = self.isBalanced(root_node.right)

        left_height = self.height
        right_height = self.height

        if abs(left_height - right_height) <=1:
            return left and right

        return False

    def calHight(self, root_node):
        if root_node is None:
            return 0
        return max(self.calHight(root_node.left), self.calHight(root_node.right)) + 1

<filename>is_balanced.py
"""
Given a Binary Tree (BT), convert it to a Doubly Linked List(DLL) In-Place. The left and right pointers in nodes are to be used as previous 
and next pointers respectively in converted DLL. The order of nodes in DLL must be same as Inorder of the given Binary Tree. The first node 
of Inorder traversal (leftmost node in BT) must be the head node of the DLL.

TreeToList

Example 1:

Input:
      1
    /  \
   3    2
Output:
3 1 2 
2 1 3 
Explanation: DLL would be 3<=>1<=>2
Example 2:

Input:
       10
      /   \
     20   30
   /   \
  40   60
Output:
40 20 60 10 30 
30 10 60 20 40
Explanation:  DLL would be 
40<=>20<=>60<=>10<=>30.
Your Task:
You don't have to take input. Complete the function bToDLL() that takes root node of the tree as a parameter and returns the head of DLL. 
The driver code prints the DLL both ways.

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(H).
Note: H is the height of the tree and this space is used implicitly for recursion stack.
"""
from math import abs, max


class Solution:
    def __init__(self) -> None:
        self.height = 0

    def isBalanced(self, root_node):
        if root_node is None:
            return True

        left = self.isBalanced(root_node.left)
        right = self.isBalanced(root_node.right)

        left_height = self.height
        right_height = self.height

        if abs(left_height - right_height) <=1:
            return left and right

        return False

    def calHight(self, root_node):
        if root_node is None:
            return 0
        return max(self.calHight(root_node.left), self.calHight(root_node.right)) + 1

Given a Binary Tree (BT), convert it to a Doubly Linked List(DLL) In-Place. The left and right pointers in nodes are to be used as previous  and next pointers respectively in converted DLL. The order of nodes in DLL must be same as Inorder of the given Binary Tree. The first node  of Inorder traversal (leftmost node in BT) must be the head node of the DLL.  TreeToList  Example 1:  Input:       1     /  \    3    2 Output: 3 1 2  2 1 3  Explanation: DLL would be 3<=>1<=>2 Example 2:  Input:        10       /   \      20   30    /   \   40   60 Output: 40 20 60 10 30  30 10 60 20 40 Explanation:  DLL would be  40<=>20<=>60<=>10<=>30. Your Task: You don't have to take input. Complete the function bToDLL() that takes root node of the tree as a parameter and returns the head of DLL.  The driver code prints the DLL both ways.  Expected Time Complexity: O(N). Expected Auxiliary Space: O(H). Note: H is the height of the tree and this space is used implicitly for recursion stack.