Leetcode 450 Delete Node in a BST

Problem:

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).

Ideas:

two steps for this problem:

1. find the node
2. If find, discuss how to delete:
   • no child: return null
   • one child: return child
   • two child return the successor

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class Solution1 {
	public TreeNode deleteNode(TreeNode root, int key) {
		if (root == null) return null;
		
		if (root.val < key) {
			// check the right node
			root.right = deleteNode(root.right, key);
		}else if (root.val > key) {
			// check the left node
			root.left = deleteNode(root.left, key);
		}else {
			// left node and right node are null
			if (root.left == null && root.right == null) return null;
			// left node is null
			if (root.left == null) return root.right;
			// right node is null
			if (root.right == null) return root.left;
			
			// left and right node are both not null
			TreeNode minNode = findMin(root.right);
			root.val = minNode.val;
			root.right = deleteNode(root.right, root.val);			
		}
		return root;
	}
	
	public TreeNode findMin(TreeNode root) {
		while(root.left != null) {
			root = root.left;	
		}
		return root;
	
	}	
}