LeetCode 450. Delete Node in a BST (javascript)

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

Search for a node to remove.
If the node is found, delete the node.

Follow up: Can you solve it with time complexity O(height of tree)?

Example 1:

Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

Example 2:

Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.

Example 3:

Input: root = [], key = 0
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 10⁴].
-10⁵ <= Node.val <= 10⁵
Each node has a unique value.
root is a valid binary search tree.
-10⁵ <= key <= 10⁵

Solution: (Recursion)

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} key
 * @return {TreeNode}
 */
var deleteNode = function(root, key) {
    if (root === null) return root;
    if (key > root.val) {
        root.right = deleteNode(root.right, key);
    } else if (key < root.val) {
        root.left = deleteNode(root.left, key);
    } else { // found the key
        // has children
        if (root.left !== null && root.right !== null) {
            let min = new TreeNode();
            min = root.right;
            while (min.left !== null) min = min.left;
            root.val = min.val;
            root.right = deleteNode(root.right, min.val);
        } else { // no children
            let newRoot = new TreeNode();
            newRoot = root.left === null ? root.right : root.left;
            root.left = root.right = null;
            return newRoot;
        }
    }
    return root;
};