#### 题目

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).

``````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

5
/ \
2   6
\   \
4   7
``````

#### 思路

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     public var val: Int
*     public var left: TreeNode?
*     public var right: TreeNode?
*     public init(_ val: Int) {
*         self.val = val
*         self.left = nil
*         self.right = nil
*     }
* }
*/
class Solution {
func deleteNode(_ root: TreeNode?, _ key: Int) -> TreeNode? {
guard let root = root else {
return nil
}

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 == nil {
return root.right
} else if root.right == nil {
return root.left
} else {
let temp = root

let newRoot = findMin(root.right!)

newRoot.right = deleteNode(root.right, newRoot.val)
newRoot.left = temp.left
return newRoot
}
}

return root
}

func findMin(_ root: TreeNode) -> TreeNode {
var _root = root
while _root.left != nil {
_root = _root.left!
}

return _root
}
}
``````