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

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode deleteNode(TreeNode root, int key) {
12         if (root == null) return null;
13         if (key < root.val) {
14             root.left = deleteNode(root.left, key);
15         }
16         else if (key > root.val) {
17             root.right = deleteNode(root.right, key);
18         }
19         else { //k == root.val
20             if (root.left == null) return root.right;
21             else if (root.right == null) return root.left;
22             else { // both root.left and root.right are not null
23                 TreeNode minRight = findMin(root.right);
24                 root.val = minRight.val;
25                 root.right = deleteNode(root.right, minRight.val);
26             }
27         }
28         return root;
29     }
30     
31     public TreeNode findMin(TreeNode cur) {
32         while (cur.left != null) {
33             cur = cur.left;
34         }
35         return cur;
36     }
37 }