# 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

### Solutions

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