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leetcode-104 Maximum Depth of Binary Tree

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    Gene Zhang
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[104] Maximum Depth of Binary Tree

Key Concept: Tree Depth Calculation - Use recursion (DFS) or level-order traversal (BFS) to find the maximum depth. Depth is 1 + max depth of subtrees.

DFS approach: Recursive bottom-up calculation. BFS approach: Count levels during level-order traversal.

# Given the root of a binary tree, return its maximum depth.
# A binary tree's maximum depth is the number of nodes along the longest path
# from the root node down to the farthest leaf node.
#
# Example 1:
# Input: root = [3,9,20,null,null,15,7]
# Output: 3
#
# Example 2:
# Input: root = [1,null,2]
# Output: 2
#
# Constraints:
# The number of nodes in the tree is in the range [0, 10^4].
# -100 <= Node.val <= 100

# 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:
    # Approach 1: DFS Recursive (Most Concise)
    def maxDepth(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

    # Approach 2: BFS Level-Order Traversal
    def maxDepth2(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0

        from collections import deque
        queue = deque([root])
        depth = 0

        while queue:
            depth += 1
            # Process all nodes at current level
            level_size = len(queue)
            for _ in range(level_size):
                node = queue.popleft()
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

        return depth

    # Approach 3: DFS Iterative with Stack
    def maxDepth3(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0

        stack = [(root, 1)]
        max_depth = 0

        while stack:
            node, depth = stack.pop()
            max_depth = max(max_depth, depth)

            if node.left:
                stack.append((node.left, depth + 1))
            if node.right:
                stack.append((node.right, depth + 1))

        return max_depth

# All approaches: Time O(n), Space O(h) where h is height