# [LeetCode] 450. Delete Node in a BST

## Problem

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.
Note: Time complexity should be O(height of tree).

Example:

``````root = [5,3,6,2,4,null,7]
key = 3

5
/ \
3   6
/ \   \
2   4   7

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 following BST.

5
/ \
4   6
/     \
2       7

5
/ \
2   6
\   \
4   7
``````

## Solution

``````class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null) return root;
if (root.val < key) root.right = deleteNode(root.right, key);
else if (root.val > key) root.left = deleteNode(root.left, key);
else {
if (root.left == null) return root.right;
if (root.right == null) return root.left;
int min = findMin(root.right);
root.val = min;
root.right = deleteNode(root.right, min);
}
return root;
}
private int findMin(TreeNode node) {
while (node.left != null) node = node.left;
return node.val;
}
}
``````