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:

1. Search for a node to remove.
2. 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

Recursively find the node that has the same value as the key, while setting the left/right nodes equal to the returned subtree
Once the node is found, have to handle the below 4 cases

node doesn‘t have left or right - return null
node only has left subtree- return the left subtree
node only has right subtree- return the right subtree
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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 /**
  * Recursively find the node that has the same value as the key, while setting the left/right nodes equal to the returned subtree
    Once the node is found, have to handle the below 4 cases

    node doesn‘t have left or right - return null
    node only has left subtree- return the left subtree
    node only has right subtree- return the right subtree
    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 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{  //equal
            if(root.left == null &&  root.right == null) return null;
            else if(root.left == null) return root.right;
            else if(root.right == null) return root.left;
            else{
                TreeNode minRight = getMin(root.right);
                root.val = minRight.val;
                root.right = deleteNode(root.right, minRight.val);
            }

        }
        return root;
    }
    public TreeNode getMin(TreeNode root){
        while(root.left != null){
            root = root.left;
        }
        return root;
    }
}