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

class Solution {
     public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null) {
            return root;
        }
        
        //find the target
        if(root.val < key) {
            root.right = deleteNode(root.right, key);
            return root;
        } else if(root.val > key) {
            root.left = deleteNode(root.left, key);
            return root;
        }
        
        //No child or only one child
        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;
        }
        
        //Two children
        if(root.right.left == null) {
            root.right.left = root.left;
            return root.right;
        } else {
            TreeNode smallest = deleteSmallest(root.right);
            smallest.left = root.left;
            smallest.right = root.right;
            return smallest;
        }
    }
    
    private TreeNode deleteSmallest(TreeNode root) {
        TreeNode cur = root.left;
        TreeNode pre = root;
        while(cur.left != null) {
            pre = cur;
            cur = cur.left;
        }
        pre.left = cur.right;
        return cur;
    }
}