Leetcode #1993: Operations on Tree
In this guide, we solve Leetcode #1993 Operations on Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Design, Array, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.