## DEV Community

Abhishek Chaudhary

Posted on

# 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.

Example 1:

Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: 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's also accepted.

Example 2:

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

Example 3:

Input: root = [], key = 0
Output: []

Constraints:

• The number of nodes in the tree is in the range `[0, 104]`.
• `-105 <= Node.val <= 105`
• Each node has a unique value.
• `root` is a valid binary search tree.
• `-105 <= key <= 105`

Follow up: Could you solve it with time complexity `O(height of tree)`?

SOLUTION:

``````# 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 insert(self, root, node):
if root:
if node:
if node.val <= root.val:
root.left = self.insert(root.left, node)
else:
root.right = self.insert(root.right, node)
else:
root = node
return root

def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
if root:
if root.val == key:
root = self.insert(root.left, root.right)
elif key < root.val:
root.left = self.deleteNode(root.left, key)
else:
root.right = self.deleteNode(root.right, key)
return root
else:
return None
``````