## The challenge

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example 1:

```Input: root = [3,9,20,null,null,15,7]
Output: 2```

Example 2:

```Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5```

Constraints:

• The number of nodes in the tree is in the range `[0, 10<sup>5</sup>]`.
• `-1000 <= Node.val <= 1000`

## The definition for a binary tree node

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 `````` `````` public class TreeNode { int val; TreeNode left; TreeNode right; TreeNode() {} TreeNode(int val) { this.val = val; } TreeNode(int val, TreeNode left, TreeNode right) { this.val = val; this.left = left; this.right = right; } } ``````

## The solution in Java code

First we need to make sure that we catch our 2 edge cases:

• root is null
• root has no child nodes

Then we can check if left is null, it means that we can run the depth again, but with right, adding 1 level.

The same then applies to the right node.

Once this recursion is out, we make sure to use the built-in `Math.min()` to do the same logic if not caught above:

• minimum value will be the left node under recursion
• compared to the right node under recursion

Making sure to increment our final output by 1, as we have not yet counted in the additional level yet.

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 `````` ``````class Solution { public int minDepth(TreeNode root) { if (root == null) return 0; if (root.left == null && root.right == null) return 1; if (root.left == null) { return this.minDepth(root.right) + 1; } if (root.right == null) { return this.minDepth(root.left) + 1; } return Math.min(this.minDepth(root.left), this.minDepth(root.right)) +1; } } ``````
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