Leetcode: 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

recursively find the node that needs to be deleted

Once the node is found, have to handle the below 4 cases

• node doesn't have left nor 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; }
  * }
  */
 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 { //k == root.val
             if (root.left == null) return root.right;
             else if (root.right == null) return root.left;
             else { // both root.left and root.right are not null
                 TreeNode minRight = findMin(root.right);
                 root.val = minRight.val;
                 root.right = deleteNode(root.right, minRight.val);
             }
         }
         return root;
     }

     public TreeNode findMin(TreeNode cur) {
         while (cur.left != null) {
             cur = cur.left;
         }
         return cur;
     }
 }