Delete Node in a BST

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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.

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: []

Solution

Way 1

/**
 * 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->left && !root->right) {
                delete root;
                return nullptr;
            } else if (!root->left) {
                auto res = root->right;
                delete root;
                return res;
            } else if (!root->right) {
                auto res = root->left;
                delete root;
                return res;
            } else { // root->left != nullptr && root->right != nullptr
                auto curr = root->right;
                while (curr->left) {
                    curr = curr->left;
                }
                curr->left = root->left;

                auto temp = root;
                root = root->right;
                delete temp;
                return root;
            }
        }

        if (root->val > key) root->left = deleteNode(root->left, key);
        if (root->val < key) root->right = deleteNode(root->right, key);

        return root;
    }
};

Way 2

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return nullptr;
        if (root->val > key) {
            root->left = deleteNode(root->left, key);
        } else if (root->val < key) {
            root->right = deleteNode(root->right, key);
        } else if (!root->left) {
            root = root->right;
        } else if (!root->right) {
            root = root->left;
        } else {
            auto successor = root->right;
            while (successor->left) {
                successor = successor->left;
            }
            root->val = successor->val;
            root->right = deleteNode(root->right, successor->val);
        }
        return root;
    }
};

Code without memory leaks:

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        // Base case: If the root is null, the tree does not contain the key.
        if (!root) return nullptr;

        // If the key to be deleted is smaller than the root's key,
        // then it lies in the left subtree.
        if (root->val > key) {
            root->left = deleteNode(root->left, key);
        }
        // If the key to be deleted is greater than the root's key,
        // then it lies in the right subtree.
        else if (root->val < key) {
            root->right = deleteNode(root->right, key);
        }
        // If the key is the same as the root's key, then this is the node
        // to be deleted.
        else {
            // Node with only one child or no child.
            if (!root->left) {
                TreeNode* temp = root->right;
                delete root; // Prevent memory leak by deallocating memory of the node to be deleted.
                return temp;
            } else if (!root->right) {
                TreeNode* temp = root->left;
                delete root; // Prevent memory leak by deallocating memory of the node to be deleted.
                return temp;
            }

            // Node with two children: Get the inorder successor (smallest
            // in the right subtree).
            TreeNode* successor = findMin(root->right);

            // Copy the inorder successor's content to this node.
            root->val = successor->val;

            // Delete the inorder successor.
            root->right = deleteNode(root->right, successor->val);
        }
        return root;
    }

private:
    // Utility function to find minimum value node in the given subtree.
    TreeNode* findMin(TreeNode* node) {
        TreeNode* current = node;
        while (current && current->left) {
            current = current->left;
        }
        return current;
    }
};

• Time Complexity: In the best case, which is a balanced BST, the time complexity is \(O(\log n)\), where \(n\) is the number of nodes. This is because, in a balanced BST, the height of the tree is \(\log n\), and in the worst case, you might need to traverse from the root to a leaf node. In the worst case, which is a skewed BST (e.g., each node has only one child), the time complexity degrades to \(O(n)\), where \(n\) is the number of nodes.
• Space Complexity: The space complexity is \(O(h)\), where \(h\) is the height of the tree. This space is used by the recursive call stack. In the best case of a balanced tree, the space complexity is \(O(\log n)\). In the worst case of a skewed tree, the space complexity becomes \(O(n)\).