Delete Node in a BST.java

/*
    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).

	Link: https://leetcode.com/problems/delete-node-in-a-bst/

    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
    
    Solution: None

    Source: https://discuss.leetcode.com/topic/65792/recursive-easy-to-understand-java-solution/2
*/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) {
            return null;
        }
        if (key < root.val) {
            root.left = deleteNode(root.left, key);
        } else if (key > root.val) {
            root.right = deleteNode(root.right, key);
        } else {
            if (root.left == null){
                return root.right;
            } else if (root.right == null) {
                return root.left;
            }
            
            TreeNode minNode = findMin(root.right);
            root.val = minNode.val;
            root.right = deleteNode(root.right, root.val);
        }
        return root;
    }
    
    private TreeNode findMin(TreeNode node){
        while (node.left != null) {
            node = node.left;
        }
        return node;
    }
}