[LeetCode] 450. Delete Node in a 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   7Given 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       7Another valid answer is [5,2,6,null,4,null,7].    5   / \  2   6   \   \    4   7

    TreeNode* deleteNode(TreeNode* root, int key) {        if (root == nullptr) return nullptr;        if (key < root->val) {            root->left = deleteNode(root->left, key);        } else if (key > root->val) {            root->right = deleteNode(root->right, key);        } else {            if (root->right) {                TreeNode **pparent = &root->right;                TreeNode *ceil = root->right;                while (ceil->left) {                    pparent = &ceil->left;                    ceil = ceil->left;                }                root->val = ceil->val;                *pparent = ceil->right;            } else if (root->left){                TreeNode **pparent = &root->left;                TreeNode *ceil = root->left;                while (ceil->right) {                    pparent = &ceil->right;                    ceil = ceil->right;                }                root->val = ceil->val;                *pparent = ceil->left;            } else {                root = nullptr;            }        }        return root;    }