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

Java代码：
/**
 * 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(root.val>key)root
4000
.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;
      root.val=findmin(root);
       }
       return root;
    }

public int findmin(TreeNode root){
TreeNode pre=root;
TreeNode nex=root.right;
while(nex.left!=null){
pre=nex;
nex=nex.left;
}
if(root==pre)pre.right=nex.right;
else pre.left=nex.right;
return nex.val;
}
}

空间复杂度:O(N)
时间复杂度:O(N)