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).

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

time = O(logn)
space = O(logn) for the BST
*/


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