450. Delete Node in a BST

1. Description

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.

2. Example

Example 1

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.

Example 2

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

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

3. 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$^5$

4. Solutions

Iteration

n is the number of nodes in root
Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    TreeNode *deleteNode(TreeNode *root, int key) {
        TreeNode dummy(0, root, nullptr);
        TreeNode **iter = &(dummy.left);

        while (*iter != nullptr) {
            if ((*iter)->val < key) {
                iter = &((*iter)->right);
            } else if ((*iter)->val == key) {
                break;
            } else {
                iter = &((*iter)->left);
            }
        }

        if (*iter == nullptr) {
            return dummy.left;
        }

        if ((*iter)->left == nullptr) {
            *iter = (*iter)->right;
        } else if ((*iter)->right == nullptr) {
            *iter = (*iter)->left;
        } else {
            TreeNode **next_iter = &((*iter)->right);
            while ((*next_iter)->left != nullptr) {
                next_iter = &((*next_iter)->left);
            }

            (*iter)->val = (*next_iter)->val;
            *next_iter = (*next_iter)->right;
        }

        return dummy.left;
    }
};