450. Delete Node in a BST

450. Delete Node in a BST (Medium)

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.

Note: Time complexity should be O(height of tree).

Example:

root = [5,3,6,2,4,null,7]
key = 3
5

/ 

3   6
/ \   

2   4   7
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 following BST.
5

/ 

4   6
/     

2       7
Another valid answer is [5,2,6,null,4,null,7].
5

/ 

2   6
\   

4   7

Companies:
Microsoft, Google, Amazon

Related Topics:
Tree

Solution 1.

// OJ: https://leetcode.com/problems/delete-node-in-a-bst/
// Author: github.com/lzl124631x
// Time: O(H)
// Space: O(H)
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return NULL;
        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) {
            auto node = root->left;
            while (node->right) node = node->right;
            root->val = node->val;
            root->left = deleteNode(root->left, root->val);
        } else if (root->right) {
            auto node = root->right;
            while (node->left) node = node->left;
            root->val = node->val;
            root->right = deleteNode(root->right, root->val);
        } else {
            delete root;
            root = NULL;
        }
        return root;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/delete-node-in-a-bst/
// Author: github.com/lzl124631x
// Time: O(H)
// Space: O(H)
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return NULL;
        if (key < root->val) {
            root->left = deleteNode(root->left, key);
        } else if (key > root->val) {
            root->right = deleteNode(root->right, key);
        } else if (!root->left) {
            auto right = root->right;
            delete root;
            return right;
        } else if (!root->right) {
            auto left = root->left;
            delete root;
            return left;
        } else {
            auto node = root->right;
            while (node->left) node = node->left;
            root->val = node->val;
            root->right = deleteNode(root->right, root->val);
        }
        return root;
    }
};