## # 450. 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   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
``````

### # Solution

Approach 1: iterative.

Approach 2: recursive.

### # Code (Python)

Approach 1:

``````    def deleteNode(self, root, key):
# iterative
node = root
parent = TreeNode(0)
parent.right = root
anchor = parent
# search for the node
while node:
if node.val == key:
break
elif node.val < key:
parent = node
node = node.right
else:
parent = node
node = node.left
if not node:
return root
# node is a leaf -- just delete the node
if not node.left and not node.right:
if parent.left == node:
parent.left = None
else:
parent.right = None
# node has one child -- move its subtree up
elif not node.left or not node.right:
if parent.left == node:
parent.left = node.left if node.left else node.right
else:
parent.right = node.left if node.left else node.right
# node has 2 children -- replace with inorder successor and delete successor
else:
# find inorder successor
successor = node.right
successor_parent = node
while successor.left:
successor_parent = successor
successor = successor.left
# exchange the values
node.val = successor.val
successor.val = key
# delete the successor --be careful that successor might have a right child
if successor_parent.right == successor:
successor_parent.right = successor.right
else:
successor_parent.left = successor.right
return anchor.right
``````

Approach 2:

``````    def deleteNode(self, root, key):
# recursive
if not root:
return None
if root.val > key:
root.left = self.deleteNode(root.left, key)
return root
elif root.val < key:
root.right = self.deleteNode(root.right, key)
return root
else:
# this happens when root.val == key, need to delete the root
if not root.left:
return root.right
if not root.right:
return root.left
# find successor
node = root.right
while node.left:
node = node.left
# the successor takes over the root's left subtree, and the new root is now root.right
node.left = root.left
return root.right
``````

Approach 1:

Approach 2: