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.
Note: Time complexity should be O(height of tree).
Example:
root = [5,3,6,2,4,null,7] key = 35
/
3 6 / \
2 4 7Given 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 7Another valid answer is [5,2,6,null,4,null,7].
5
/
2 6 \
4 7
Companies:
Microsoft, Google, Amazon
Related Topics:
Tree
// 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 p = root->left;
while (p->right) p = p->right;
root->val = p->val;
root->left = deleteNode(root->left, root->val);
} else if (root->right) {
auto p = root->right;
while (p->left) p = p->left;
root->val = p->val;
root->right = deleteNode(root->right, root->val);
} else {
delete root;
root = NULL;
}
return root;
}
};
// 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 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;
}
};