1993. Operations on Tree

解題說明

C++ 解法

複雜度分析

虛擬碼

1. Constructor:
   a. Build children list from parent array
   b. Initialize lockedBy array with -1 (unlocked)
2. lock(num, user):
   a. If lockedBy[num] != -1: return false
   b. Set lockedBy[num] = user, return true
3. unlock(num, user):
   a. If lockedBy[num] != user: return false
   b. Set lockedBy[num] = -1, return true
4. upgrade(num, user):
   a. If lockedBy[num] != -1: return false
   b. Walk up ancestors; if any locked: return false
   c. DFS descendants: unlock all locked ones, track if any found
   d. If no locked descendant found: return false
   e. Set lockedBy[num] = user, return true

其他解法

延伸思考