# LeetCode 450. Delete Node in a BST

2017-01-04

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

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.

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``` ```/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null) return null;
if (key < root.val) {
root.left = deleteNode(root.left, key);
return root;
}
if (key > root.val) {
root.right = deleteNode(root.right, key);
return root;
}
if (root.left != null) {
TreeNode left = root.left;
if (left.right == null) {
left.right = root.right;
return left;
}
while (left.right != null && left.right.right != null) {
left = left.right;
}
if (left.right.left == null) {
root.val = left.right.val;
left.right = null;
return root;
}
root.val = left.right.val;
left.right = left.right.left;
return root;
} else {
return root.right;
}
}
}```