LeetCode-450. Delete Node in a BST [C++][Java]

LeetCode-450. Delete Node in a BSTLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/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.

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, 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

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

【C++】

/**
 * 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 nullptr;}
        if (root->val == key) {
            if(root->left == nullptr && root->right == nullptr)  {
                delete root;
                return nullptr;
            }
            if (root->left && root->right == nullptr) {
                TreeNode* temp = root->left;
                delete root;
                return temp;
            }
            if (root->left == nullptr && root->right) {
                TreeNode* temp = root->right;
                delete root;
                return temp;
            }
            if(root->left && root->right) {
                TreeNode* cur = root->right;
                while (cur->left != nullptr) {cur = cur->left;}
                int newval = cur->val;
                root->val = newval;
                root->right = deleteNode(root->right, newval);
                return root;
            }
        }
        if (root->val > key) {root->left = deleteNode(root->left, key);}
        else {root->right = deleteNode(root->right, key);}
        return root;
    }
};

【Java】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) {return null;}
        if (root.val == key) {
            if(root.left == null && root.right == null)  {
                return null;
            }
            if (root.left != null && root.right == null) {
                return root.left;
            }
            if (root.left == null && root.right != null) {
                return root.right;
            }
            if (root.left != null && root.right != null) {
                TreeNode cur = root.right;
                while (cur.left != null) {cur = cur.left;}
                int newval = cur.val;
                root.val = newval;
                root.right = deleteNode(root.right, newval);
                return root;
            }
        }
        if (root.val > key) {root.left = deleteNode(root.left, key);}
        else {root.right = deleteNode(root.right, key);}
        return root;
    }
}