How do you solve Operations on Tree on LeetCode?

Operations on Tree (LeetCode #1993) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding tree traversal is crucial for managing parent-child relationships.

What is the brute force solution for Operations on Tree?

The brute force approach for Operations on Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each condition for locking, unlocking, and upgrading nodes by traversing the tree. This is straightforward but inefficient, as we may need to check multiple nodes repeatedly. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Operations on Tree?

Clearly explain your thought process while coding. Discuss edge cases and how your solution handles them. Practice tree traversal techniques to become comfortable with tree problems.

What are common mistakes when solving Operations on Tree?

Not checking all conditions for upgrade correctly. Assuming a node can be locked without checking if it's already locked.

#1993

Operations on Tree

Medium

ArrayHash TableTreeDepth-First SearchBreadth-First SearchDesignHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

This approach efficiently manages the locking and unlocking operations by using a combination of a parent-child relationship and a map to track locked nodes. It allows us to quickly check conditions for locking, unlocking, and upgrading without redundant traversals.

⚙️

Algorithm

4 steps

1Step 1: Store the parent-child relationships in a list and maintain a map for locked nodes.
2Step 2: For locking, check if the node is unlocked and lock it for the user.
3Step 3: For unlocking, check if the node is locked by the same user and unlock it.
4Step 4: For upgrading, check the three conditions (unlocked, has locked descendant, no locked ancestor) and perform the operations accordingly.

solution.py51 lines

1class LockingTree:
2    def __init__(self, parent):
3        self.parent = parent
4        self.locked = {}
5        self.children = [[] for _ in range(len(parent))]
6        for i in range(1, len(parent)):
7            self.children[parent[i]].append(i)
8
9    def lock(self, num, user):
10        if num in self.locked:
11            return False
12        self.locked[num] = user
13        return True
14
15    def unlock(self, num, user):
16        if num not in self.locked or self.locked[num] != user:
17            return False
18        del self.locked[num]
19        return True
20
21    def upgrade(self, num, user):
22        if num in self.locked:
23            return False
24        if not self.has_locked_descendant(num):
25            return False
26        if self.has_locked_ancestor(num):
27            return False
28        self.locked[num] = user
29        self.unlock_descendants(num)
30        return True
31
32    def has_locked_descendant(self, num):
33        if num in self.locked:
34            return True
35        for child in self.children[num]:
36            if self.has_locked_descendant(child):
37                return True
38        return False
39
40    def has_locked_ancestor(self, num):
41        while num != -1:
42            if num in self.locked:
43                return True
44            num = self.parent[num]
45        return False
46
47    def unlock_descendants(self, num):
48        if num in self.locked:
49            del self.locked[num]
50        for child in self.children[num]:
51            self.unlock_descendants(child)

ℹ

Complexity note: This complexity is due to the fact that we can efficiently check for locked nodes and their relationships using a map and a list, allowing us to avoid redundant traversals.

1Understanding tree traversal is crucial for managing parent-child relationships.
2Efficiently tracking locked nodes can significantly reduce operation time.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.