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Kotlin-based LeetCode algorithm problem solutions, regularly updated
1993\. Operations on Tree
Medium
You are given a tree with `n` nodes numbered from `0` to `n - 1` in the form of a parent array `parent` where `parent[i]` is the parent of the ith
node. The root of the tree is node `0`, so `parent[0] = -1` since it has no parent. You want to design a data structure that allows users to lock, unlock, and upgrade nodes in the tree.
The data structure should support the following functions:
* **Lock: Locks** the given node for the given user and prevents other users from locking the same node. You may only lock a node using this function if the node is unlocked.
* **Unlock: Unlocks** the given node for the given user. You may only unlock a node using this function if it is currently locked by the same user.
* **Upgrade: Locks** the given node for the given user and **unlocks** all of its descendants **regardless** of who locked it. You may only upgrade a node if **all** 3 conditions are true:
* The node is unlocked,
* It has at least one locked descendant (by **any** user), and
* It does not have any locked ancestors.
Implement the `LockingTree` class:
* `LockingTree(int[] parent)` initializes the data structure with the parent array.
* `lock(int num, int user)` returns `true` if it is possible for the user with id `user` to lock the node `num`, or `false` otherwise. If it is possible, the node `num` will become **locked** by the user with id `user`.
* `unlock(int num, int user)` returns `true` if it is possible for the user with id `user` to unlock the node `num`, or `false` otherwise. If it is possible, the node `num` will become **unlocked**.
* `upgrade(int num, int user)` returns `true` if it is possible for the user with id `user` to upgrade the node `num`, or `false` otherwise. If it is possible, the node `num` will be **upgraded**.
**Example 1:**
![](https://assets.leetcode.com/uploads/2021/07/29/untitled.png)
**Input**
["LockingTree", "lock", "unlock", "unlock", "lock", "upgrade", "lock"]
[[[-1, 0, 0, 1, 1, 2, 2]], [2, 2], [2, 3], [2, 2], [4, 5], [0, 1], [0, 1]]
**Output:** [null, true, false, true, true, true, false]
**Explanation:**
LockingTree lockingTree = new LockingTree([-1, 0, 0, 1, 1, 2, 2]);
lockingTree.lock(2, 2); // return true because node 2 is unlocked.
// Node 2 will now be locked by user 2.
lockingTree.unlock(2, 3); // return false because user 3 cannot unlock a node locked by user 2.
lockingTree.unlock(2, 2); // return true because node 2 was previously locked by user 2.
// Node 2 will now be unlocked.
lockingTree.lock(4, 5); // return true because node 4 is unlocked.
// Node 4 will now be locked by user 5.
lockingTree.upgrade(0, 1); // return true because node 0 is unlocked and has at least one locked descendant (node 4).
// Node 0 will now be locked by user 1 and node 4 will now be unlocked.
lockingTree.lock(0, 1); // return false because node 0 is already locked.
**Constraints:**
* `n == parent.length`
* `2 <= n <= 2000`
* `0 <= parent[i] <= n - 1` for `i != 0`
* `parent[0] == -1`
* `0 <= num <= n - 1`
* 1 <= user <= 104
* `parent` represents a valid tree.
* At most `2000` calls **in total** will be made to `lock`, `unlock`, and `upgrade`.