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Maximum Depth of Binary Tree

Easy
DFS

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 Tree:

    3
   / \
  9  20
    /  \
   15   7

Maximum depth: 3 (path: 3 -> 20 -> 15 or 3 -> 20 -> 7)

This is a classic DFS problem where we explore each path to its deepest point and track the maximum depth encountered.

Approach:

  • Base case: If node is null, depth is 0
  • Recursive case: depth = 1 + max(leftDepth, rightDepth)

Example 1

Input: root = [3,9,20,null,null,15,7]
Output: 3
Explanation: The deepest path is 3 -> 20 -> 15 (or 3 -> 20 -> 7), which has 3 nodes.

Example 2

Input: root = [1,null,2]
Output: 2
Explanation: The tree has two levels: root (1) and its right child (2).

Example 3

Input: root = []
Output: 0
Explanation: An empty tree has depth 0.

Constraints

  • •The number of nodes in the tree is in the range [0, 10^4].
  • •-100 <= Node.val <= 100
Show Hints (4)
Hint 1: What is the depth of an empty tree? What is the depth of a single node?
Hint 2: The depth at any node is 1 plus the maximum depth of its subtrees.
Hint 3: DFS naturally explores depth-first, making this problem straightforward.
Hint 4: Can you also solve this with BFS by counting levels?
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