450. Delete Node in a BST

2023, Mar 23 2 mins read

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:

Search for a node to remove.
If the node is found, delete the node.

2. Example

Example 1

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 1

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 parent_node = TreeNode();
        parent_node.left = root;

        TreeNode **link_to_node = &parent_node.left, *target_node = root;
        while (target_node != nullptr && target_node->val != key) {
            if (target_node->val < key) {
                link_to_node = &target_node->right;
                target_node = target_node->right;
            } else if (target_node->val > key) {
                link_to_node = &target_node->left;
                target_node = target_node->left;
            }
        }

        if (target_node != nullptr) {
            if (target_node->left == nullptr && target_node->right == nullptr) {
                *link_to_node = nullptr;
            } else if (target_node->left == nullptr && target_node->right != nullptr) {
                *link_to_node = target_node->right;
            } else if (target_node->left != nullptr && target_node->right == nullptr) {
                *link_to_node = target_node->left;
            } else if (target_node->left != nullptr && target_node->right != nullptr) {
                TreeNode **link_to_next = &target_node->right, *next_node = target_node->right;
                while (next_node->left != nullptr) {
                    link_to_next = &next_node->left;
                    next_node = next_node->left;
                }

                target_node->val = next_node->val;
                *link_to_next = next_node->right;
            }
        }

        return parent_node.left;
    }
};