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LeetCode 450 Delete Node In ABST - Medium

450. Delete Node in a BST -- Medium

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450. Delete Node In ABST — Medium

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Problem

  1. Delete Node in a BST -- Medium

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

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

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 deleteNode(self, root: TreeNode, key: int) -> TreeNode:
        if not root:
            return
        if key > root.val:
            root.right = self.deleteNode(root.right, key)
        elif key < root.val:
            root.left = self.deleteNode(root.left, key)
        # now the key is the root of a subtree
        else:
 
            if not root.left:
                return root.right
            # if it has a left child, we want to find the max val on the left subtree to 
            # replace the node we want to delete.
            else:
                # try to find the max value on the left subtree
                tmp = root.left
                while tmp.right:
                    tmp = tmp.right
                # replace
                root.val = tmp.val
                root.left = self.deleteNode(root.left, tmp.val)
        return root

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