"""
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
"""

class Solution(object):
    def delete_node(self, root, key):
        """
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        """
        if not root: return None

        if root.val == key:
            if root.left:
                # Find the right most leaf of the left sub-tree
                left_right_most = root.left
                while left_right_most.right:
                    left_right_most = left_right_most.right
                # Attach right child to the right of that leaf
                left_right_most.right = root.right
                # Return left child instead of root, a.k.a delete root
                return root.left
            else:
                return root.right
        # If left or right child got deleted, the returned root is the child of the deleted node.
        elif root.val > key:
            root.left = self.deleteNode(root.left, key)
        else:
            root.right = self.deleteNode(root.right, key)
        return root