# GeetCode Hub

Given the `root` of a binary tree and two integers `val` and `depth`, add a row of nodes with value `val` at the given depth `depth`.

Note that the `root` node is at depth `1`.

• Given the integer `depth`, for each not null tree node `cur` at the depth `depth - 1`, create two tree nodes with value `val` as `cur`'s left subtree root and right subtree root.
• `cur`'s original left subtree should be the left subtree of the new left subtree root.
• `cur`'s original right subtree should be the right subtree of the new right subtree root.
• If `depth == 1` that means there is no depth `depth - 1` at all, then create a tree node with value `val` as the new root of the whole original tree, and the original tree is the new root's left subtree.

Example 1:

```Input: root = [4,2,6,3,1,5], val = 1, depth = 2
Output: [4,1,1,2,null,null,6,3,1,5]
```

Example 2:

```Input: root = [4,2,null,3,1], val = 1, depth = 3
Output: [4,2,null,1,1,3,null,null,1]
```

Constraints:

• The number of nodes in the tree is in the range `[1, 104]`.
• The depth of the tree is in the range `[1, 104]`.
• `-100 <= Node.val <= 100`
• `-105 <= val <= 105`
• `1 <= depth <= the depth of tree + 1`

/** * 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; * } * } */ class Solution { public TreeNode addOneRow(TreeNode root, int val, int depth) { } }