LeetCode - Delete Node in a BST

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.
Note: Time complexity should be O(height of tree).

Example:

root = [5,3,6,2,4,null,7]
key = 3

    5
   /   3   6
 / \   2   4   7

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 following BST.

    5
   /   4   6
 /     2       7

Another valid answer is [5,2,6,null,4,null,7].

    5
   /   2   6
   \       4   7

/**
 * 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 deleteNode(TreeNode root, int key) {
        if (root == null) {
            return null;
        }
        if (root.val > key) {
            root.left = deleteNode(root.left, key);
        } else if (root.val < key) {
            root.right = deleteNode(root.right, key);
        } else if (root.left != null && root.right != null){
            TreeNode node = findMax(root.left);
            root.val = node.val;
            root.left = deleteNode(root.left, node.val);
        } else if (root.left != null){
            root = root.left;
        } else {
            root = root.right;
        }
        return root;
    }
    
    public TreeNode findMax(TreeNode root) {
        while(root != null && root.right != null) {
            root = root.right;
        }
        return root;
    }

}