Formatted question description: https://leetcode.ca/all/450.html
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:
- 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 = 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/
// 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;
}
};
Solution 2.
// OJ: https://leetcode.com/problems/delete-node-in-a-bst/
// 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;
}
};
Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null)
return root;
if (root.val == key) {
TreeNode dummyRoot = new TreeNode(0);
dummyRoot.left = root;
if (root.right != null) {
TreeNode successor = root.right;
if (successor.left == null) {
successor.left = root.left;
dummyRoot.left = successor;
} else {
TreeNode parent = successor, child = successor.left;
while (child.left != null) {
child = child.left;
parent = parent.left;
}
root.val = child.val;
parent.left = child.right;
}
} else
dummyRoot.left = root.left;
return dummyRoot.left;
} else {
TreeNode parent = root;
TreeNode child = root.val > key ? root.left : root.right;
while (child != null && child.val != key) {
parent = child;
child = child.val > key ? child.left : child.right;
}
if (child != null) {
boolean isLeftChild = parent.left == child;
if (child.left == null && child.right == null) {
if (isLeftChild)
parent.left = null;
else
parent.right = null;
} else if (child.left == null) {
if (isLeftChild)
parent.left = child.right;
else
parent.right = child.right;
} else if (child.right == null) {
if (isLeftChild)
parent.left = child.left;
else
parent.right = child.left;
} else {
TreeNode temp1 = child.right;
if (temp1.left == null) {
child.val = temp1.val;
child.right = temp1.right;
} else {
TreeNode temp2 = temp1.left;
while (temp2.left != null) {
temp2 = temp2.left;
temp1 = temp1.left;
}
child.val = temp2.val;
temp1.left = temp2.right;
}
}
}
return root;
}
}
}