package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

/**
 * 450. 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
 */
public class _450 {
    public static class Solution1 {

        /**
         * credit: https://discuss.leetcode.com/topic/65792/recursive-easy-to-understand-java-solution
         * Steps:
         * 1. Recursively find the node that has the same value as the key, while setting the left/right nodes equal to the returned subtree
         * 2. Once the node is found, have to handle the below 4 cases
         * a. node doesn't have left or right - return null
         * b. node only has left subtree- return the left subtree
         * c. node only has right subtree- return the right subtree
         * d. node has both left and right - find the minimum value in the right subtree, set that value to the currently found node, then recursively delete the minimum value in the right subtree
         */
        public TreeNode deleteNode(TreeNode root, int key) {
            if (root == null) {
                return root;
            }
            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 = getMin(root.right);
                root.val = minNode.val;
                root.right = deleteNode(root.right, root.val);
            }
            return root;
        }

        private TreeNode getMin(TreeNode node) {
            while (node.left != null) {
                node = node.left;
            }
            return node;
        }
    }

}

@@ -82,772 +82,8 @@
 e;%0A%0A
-/**%0A * 450. Delete Node in a BST%0A *%0A * Given a root node reference of a BST and a key, delete the node with the given key in the BST.%0A * Return the root node reference (possibly updated) of the BST.%0A%0A Basically, the deletion can be divided into two stages:%0A%0A Search for a node to remove.%0A If the node is found, delete the node.%0A Note: Time complexity should be O(height of tree).%0A%0A Example:%0A%0A root = %5B5,3,6,2,4,null,7%5D%0A key = 3%0A%0A     5%0A    / %5C%0A   3   6%0A  / %5C   %5C%0A 2  4   7%0A%0A Given key to delete is 3. So we find the node with value 3 and delete it.%0A%0A One valid answer is %5B5,4,6,2,null,null,7%5D, shown in the following BST.%0A%0A     5%0A    / %5C%0A   4   6%0A  /     %5C%0A 2       7%0A%0A Another valid answer is %5B5,2,6,null,4,null,7%5D.%0A%0A     5%0A    / %5C%0A   2   6%0A   %5C   %5C%0A   4   7%0A */%0A
 publ

refactor 450