<p>Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return <em>the <strong>root node reference</strong> (possibly updated) of the BST</em>.</p>

<p>Basically, the deletion can be divided into two stages:</p>

<ol>
	<li>Search for a node to remove.</li>
	<li>If the node is found, delete the node.</li>
</ol>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/04/del_node_1.jpg" style="width: 800px; height: 214px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,7], key = 3
<strong>Output:</strong> [5,4,6,2,null,null,7]
<strong>Explanation:</strong> 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 above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it&#39;s also accepted.
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/04/del_node_supp.jpg" style="width: 350px; height: 255px;" />
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,7], key = 0
<strong>Output:</strong> [5,3,6,2,4,null,7]
<strong>Explanation:</strong> The tree does not contain a node with value = 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [], key = 0
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
	<li>Each node has a <strong>unique</strong> value.</li>
	<li><code>root</code> is a valid binary search tree.</li>
	<li><code>-10<sup>5</sup> &lt;= key &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you solve it with time complexity <code>O(height of tree)</code>?</p>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
        if not root:
            return 
        if root.val == key:
            if not root.right:
                return root.left
            
            if not root.left:
                return root.right
            
            tmp = root.right
            while tmp.left:
                tmp = tmp.left
            
            root.val = tmp.val
            root.right = self.deleteNode(root.right, tmp.val)
        elif root.val<key:
            root.right = self.deleteNode(root.right, key)
        else:
            root.left = self.deleteNode(root.left, key)
        
        return root

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