[hard]450. Delete Node in a BST

450. 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:
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

Solution:

It reminds me of my high school days. It's a very basic step in treap, and now I just need to remove the fix from treap. Super easy.

1) If x has no sons, just delete it. don't forget x <- NULL

2) if x has a son, delete x and replace x with its son

2) if x has two son, right rotate x and deleteNode(x -> right)

(I assume you know what is a rotate operation in treap)

Here is my code:

class Solution {
public:
TreeNode* deleteNode(TreeNode*& x, int key) {
if (x) {
if (x -> val == key) {
if (x -> left == NULL && x -> right == NULL) {
delete x;
x = NULL;
} else if (x -> left == NULL) {
auto y = x -> right;
delete x; x = y;
} else if (x -> right == NULL) {
auto y = x -> left;
delete x; x = y;
} else {
auto y = x -> left;
x -> left = y -> right;
y -> right = x;
x = y;
deleteNode(x -> right, key);
}
} else if (x -> val > key)
deleteNode(x -> left, key);
else
deleteNode(x -> right, key);
}
return x;
}
};