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450. Delete Node in a BST

Difficulty Topics

Description

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

delete-node-in-a-bst.py
# 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 deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:

        def go(node, key):
            if not node: return None

            if node.val > key:
                node.left = go(node.left, key)
            elif node.val < key:
                node.right = go(node.right, key)
            else:
                if not node.left:
                    return node.right

                if not node.right:
                    return node.left

                successor = node.left
                while successor.right:
                    successor = successor.right

                node.val = successor.val
                node.left = go(node.left, node.val)

            return node

        return go(root, key)