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