Operations on Tree — LeetCode 1993 Python Solution

MediumTreeDepth-First SearchBreadth-First SearchDesignArrayHash Table
Problem
#1993
Reading time
9 min

The problem

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.

Example

Input
["LockingTree", "lock", "unlock", "unlock", "lock", "upgrade", "lock"]
Output
[null, true, false, true, true, true, false]
Explanation
LockingTree lockingTree = new LockingTree([-1, 0, 0, 1, 1, 2, 2]);

Python solution

Python
class LockingTree:
    def __init__(self, parent: List[int]):
        n = len(parent)
        self.locked = [-1] * n
        self.parent = parent
        self.children = [[] for _ in range(n)]
        for son, fa in enumerate(parent[1:], 1):
            self.children[fa].append(son)

    def lock(self, num: int, user: int) -> bool:
        if self.locked[num] == -1:
            self.locked[num] = user
            return True
        return False

    def unlock(self, num: int, user: int) -> bool:
        if self.locked[num] == user:
            self.locked[num] = -1
            return True
        return False

    def upgrade(self, num: int, user: int) -> bool:
        def dfs(x: int):
            nonlocal find
            for y in self.children[x]:
                if self.locked[y] != -1:
                    self.locked[y] = -1
                    find = True
                dfs(y)

        x = num
        while x != -1:
            if self.locked[x] != -1:
                return False
            x = self.parent[x]

        find = False
        dfs(num)
        if not find:
            return False
        self.locked[num] = user
        return True


# Your LockingTree object will be instantiated and called as such:
# obj = LockingTree(parent)
# param_1 = obj.lock(num,user)
# param_2 = obj.unlock(num,user)
# param_3 = obj.upgrade(num,user)

Complexity

MeasureComplexity
TimeO(n)
SpaceO(n) auxiliary

Pattern: Tree Traversal

Choose the order — preorder, inorder, postorder, level — and the problem solves itself. LeetCode 1993. Operations on Tree is filed here because LeetCode tags it Tree, which is the vocabulary this hub collects.

The tree traversal guide has the Python template for the pattern and the 225 LeetCode problems that use it.

Related problems

Frequently asked questions

How hard is LeetCode 1993. Operations on Tree?
LeetCode 1993. Operations on Tree is rated Medium on LeetCode.
What is the time complexity of LeetCode 1993. Operations on Tree?
The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 1993. Operations on Tree?
The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 1993. Operations on Tree cover?
LeetCode 1993. Operations on Tree is tagged Tree, Depth-First Search, Breadth-First Search, Design, Array and Hash Table on LeetCode.

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