450. Delete Node in a BST

https://leetcode.com/problems/delete-node-in-a-bst/

Problem

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.

Follow up: Can you solve it with time complexity O(height of tree)?

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: []

Constraints:

• The number of nodes in the tree is in the range [0, 104].

• -105 <= Node.val <= 105

• Each node has a unique value.

• root is a valid binary search tree.

• -105 <= key <= 105

Solution

/**
 * 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 == nullptr) return root;
        TreeNode* par = nullptr, *cur = root;
        while (cur != nullptr) {
            if (cur->val > key) {
                par = cur;
                cur = cur->left;
            } else if (cur->val < key) {
                par = cur;
                cur = cur->right;
            } else {
                break;
            }
        }
        if (cur == nullptr) return root;
        if (cur->right == nullptr) {
            cur = cur->left;
        } else {
            if (cur->right->left == nullptr) {
                cur->right->left = cur->left;
                cur = cur->right;
            } else {
                TreeNode *new_cur = cur->right;
                while (new_cur->left->left != nullptr) new_cur = new_cur->left;
                TreeNode *temp = new_cur->left;
                new_cur->left = temp->right;
                new_cur = temp;
                new_cur->right = cur->right;
                new_cur->left = cur->left;
                cur = new_cur;
            }
        }
        if (par == nullptr) return cur;
        if (cur != nullptr) {
            if (par->val > cur->val) par->left = cur;
            else par->right = cur;
        } else {
            if (par->left != nullptr && par->left->val == key) par->left = cur;
            else par->right = cur;
        }
        return root;
    }
};

• #bst

• #tree

• #important