# Delete Node In A BST Problem

## Description

LeetCode Problem 450.

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.

Example 1:

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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.
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/04/del_node_supp.jpg" style="width: 350px; height: 255px;" />
``````

Example 2:

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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:

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Input: root = [], key = 0
Output: []
``````

Constraints:

• The number of nodes in the tree is in the range [0, 10^4].
• -10^5 <= Node.val <= 10^5
• Each node has a unique value.
• root is a valid binary search tree.
• -10^5 <= key <= 10^5

## Sample C++ Code

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/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* deleteNode(TreeNode* root, int key) {
if (!root) return nullptr;
if (root->val == key) {
if (!root->right) {
TreeNode* left = root->left;
delete root;
return left;
}
else {
TreeNode* right = root->right;
while (right->left)
right = right->left;
swap(root->val, right->val);
}
}
root->left = deleteNode(root->left, key);
root->right = deleteNode(root->right, key);
return root;
}
};
``````