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Kotlin-based LeetCode algorithm problem solutions, regularly updated
450\. 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:
1. Search for a node to remove.
2. If the node is found, delete the node.
**Example 1:**
![](https://assets.leetcode.com/uploads/2020/09/04/del_node_1.jpg)
**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.
![](https://assets.leetcode.com/uploads/2020/09/04/del_node_supp.jpg)
**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)`?