MaximumDepthofBinaryTree_Day63.py

#Brute Approach
class Solution:
    def maxDepth(self, root: Optional[TreeNode]) -> int:
        if root is None:
            return 0
        left_depth = self.maxDepth(root.left)
        right_depth = self.maxDepth(root.right)
        return 1 + max(left_depth, right_depth)


#Better Approach

from collections import deque

class Solution:
    def maxDepth(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        queue = deque([root])
        depth = 0
        while queue:
            for _ in range(len(queue)):
                node = queue.popleft()
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            depth += 1
        return depth

#Optimal Approach

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxDepth(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        stack = [(root, 1)]  # (node, current_depth)
        max_depth = 0
        while stack:
            node, depth = stack.pop()
            if node:
                max_depth = max(max_depth, depth)
                stack.append((node.left, depth + 1))
                stack.append((node.right, depth + 1))
        return max_depth

# Time Complexity: O(N)
# Space Complexity: O(H), iterative stack holds at most height H.