delete-node-in-a-bst_450.py

# 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.


# 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: []

# ---------------------------------------Runtime 74 ms Beats 47.14% Memory 20.8 MB Beats 49.73%---------------------------------------


from typing import Optional


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 root

        if key > root.val:
            root.right = self.deleteNode(root.right, key)
        elif key < root.val:
            root.left = self.deleteNode(root.left, key)
        else:
            if not root.left:
                return root.right
            if not root.right:
                return root.left

            cur = root.right

            while cur.left:
                cur = cur.left
            root.val = cur.val
            root.right = self.deleteNode(root.right, root.val)
        return root