0104. Maximum Depth of Binary Tree

Given the root of a binary tree, return its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

To implement the solution for finding the maximum depth of a binary tree, we need to use a depth-first search (DFS) approach. Every time we traverse down a node, we increase our depth count. We need to check both left and right subtrees to find the maximum depth from the current node.

Here's a step-by-step plan:

1. If the current node is null, that contributes 0 to the depth.
2. Otherwise, compute the depth of the left and right subtrees.
3. The depth of the current node is 1 plus the maximum of the depths of the left and right subtrees.
4. Return this computed depth.

Time Complexity: O(N), where N is the number of nodes in the tree. We have to visit each node once. Space Complexity: O(H), where H is the height of the tree. This is due to the recursion stack.

Consider the binary tree:

``` 3 / \ 9 20 / \ 15 7 ```

The maximum depth of this tree is 3.

Here is a skeleton code for implementing the solution. An additional helper function can also be included for testing purposes using examples.

Test Case 1: Passed (Expected: 3, Got: 3)
Test Case 2: Passed (Expected: 2, Got: 2)
Test Case 3: Passed (Expected: 1, Got: 1)
Test Case 4: Passed (Expected: 0, Got: 0)
Test Case 5: Passed (Expected: 4, Got: 4)
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