LC104 Maximum Depth of Binary Tree 🟢

This matches the "bottom-up combine" pattern — both children report depth first, then you summarize at the current node.

1. Mindset Shift

The naive way walks every root-to-leaf path and takes the longest — correct, but awkward to code.

Key insight: depth(tree) = 1 + max(depth(left), depth(right)). Let the children compute their own depths (postorder), then combine.

2. The Logic — 4 Steps

1. Base case — if node is None → return 0
2. Ask left — left = maxDepth(node.left)
3. Ask right — right = maxDepth(node.right)
4. Combine — return 1 + max(left, right)

3. LeetCode-Ready Code

# 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 root is None:
            return 0
        left = self.maxDepth(root.left)
        right = self.maxDepth(root.right)
        return 1 + max(left, right)

4. Dry Run — [3,9,20,null,null,15,7]

5. Edge Cases & Pitfalls

• Forgetting the None base case → AttributeError on .left/.right
• Count nodes, not edges — always add 1 for the current node

6. Time & Space Complexity

• Time O(n) — visit every node once
• Space O(h) — call-stack depth equals tree height (worst case h = n)