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Problem solution · Python

Operations on Tree

Operations on Tree: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
52 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Operations on Tree, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 52 lines of Python from the credited upstream file 1993.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeOperations on Tree · PythonPython
Use this to learn the idea, then write your own version.
class Node:  def __init__(self):    self.children: list[int] = []    self.lockedBy = -1  class LockingTree:  def __init__(self, parent: list[int]):    self.parent = parent    self.nodes = [Node() for _ in range(len(parent))]    for i in range(1, len(parent)):      self.nodes[parent[i]].children.append(i)   def lock(self, num: int, user: int) -> bool:    if self.nodes[num].lockedBy != -1:      return False    self.nodes[num].lockedBy = user    return True   def unlock(self, num: int, user: int) -> bool:    if self.nodes[num].lockedBy != user:      return False    self.nodes[num].lockedBy = -1    return True   def upgrade(self, num: int, user: int) -> bool:    if self.nodes[num].lockedBy != -1:      return False    if not self._anyLockedDescendant(num):      return False     # Walk up the hierarchy to ensure that there are no locked ancestors.    i = num    while i != -1:      if self.nodes[i].lockedBy != -1:        return False      i = self.parent[i]     self._unlockDescendants(num)    self.nodes[num].lockedBy = user    return True   def _anyLockedDescendant(self, i: int) -> bool:    return (self.nodes[i].lockedBy != -1 or            any(self._anyLockedDescendant(child)            for child in self.nodes[i].children))   def _unlockDescendants(self, i: int) -> None:    self.nodes[i].lockedBy = -1    for child in self.nodes[i].children:      self._unlockDescendants(child)

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