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   7Given 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       7Another 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.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)