Maximum Depth of Binary Tree

Intuition

• On each subroutine, return 1 + the max of any of the two branches.

Complexity

Runtime

$O(N)$ since you traverse through every node once.

Space

$O(N)$ since the height of a binary tree can be N nodes long, meaning N nodes of stack space, with $O(logN)$ space in the case of a balanced tree, since there are $log(N)$ nodes on every level. (n / 2 / 2 / 2 …) until $\frac{n}{2^k}$ = 1.

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 not root:
            return 0
        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

Notes

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References

Maximum Depth of Binary Tree - LeetCode