# How to Count the Minimum Depth of a Binary Tree in Java ## 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, 105]`.
• `-1000 <= Node.val <= 1000`

## The definition for a binary tree node

``` 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.

```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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