will this code pass the problems staremnt all testcases and no tle analyze it perfectly 
"""
import sys

class Node:
    def __init__(self, val):
        self.val = val
        self.uid = None
        self.parent = None  
        self.child = [] 
        self.isLocked = False  
        self.locked_descendants = set()  

class Tree:
    def __init__(self, m, n, name_list):
        self.node_list = []
        self.node_map = {}
        for name in name_list:
            node = Node(name)
            self.node_list.append(node)
            self.node_map[name] = node

        # Build the tree
        for i in range(n):
            for j in range(m):
                child_index = m * i + j + 1
                if child_index < n:
                    self.node_list[i].child.append(self.node_list[child_index])
                    self.node_list[child_index].parent = self.node_list[i]

        self.root = self.node_list[0]

    def lock(self, node, uid):
        if node.isLocked:
            return False

        # Check if any ancestor is locked
        current = node.parent
        while current:
            if current.isLocked:
                return False
            current = current.parent

        # Check if any descendant is locked
        if node.locked_descendants:
            return False

        # Lock the node
        node.isLocked = True
        node.uid = uid

        # Update locked_descendants for all ancestors
        current = node.parent
        while current:
            current.locked_descendants.add(node)
            current = current.parent

        return True

    def unlock(self, node, uid):
        if not node.isLocked or node.uid != uid:
            return False

        # Unlock the node
        node.isLocked = False
        node.uid = None

        # Update locked_descendants for all ancestors
        current = node.parent
        while current:
            current.locked_descendants.discard(node)  # Correct
            current = current.parent

        return True

    def upgrade(self, node, uid):
        if node.isLocked:
            return False

        # Check if any ancestor is locked
        current = node.parent
        while current:
            if current.isLocked:
                return False
            current = current.parent

        # Check if there is at least one locked descendant
        if not node.locked_descendants:
            return False

        # Collect all locked descendants
        locked_descendants = list(node.locked_descendants)

        # Check if all locked descendants have the same uid
        for ld in locked_descendants:
            if ld.uid != uid:
                return False

        # Unlock all locked descendants
        for ld in locked_descendants:
            ld.isLocked = False
            ld.uid = None
            node.locked_descendants.discard(ld)

            # Update locked_descendants for all ancestors of ld
            ancestor = ld.parent
            while ancestor:
                ancestor.locked_descendants.discard(ld)
                ancestor = ancestor.parent

        # Lock the current node
        node.isLocked = True
        node.uid = uid

        # Update locked_descendants for all ancestors
        current = node.parent
        while current:
            current.locked_descendants.add(node)
            current = current.parent

        return True

    def process_queries(self, queries):
        results = []
        for query in queries:
            parts = query.strip().split()
            if len(parts) != 3:
                results.append('false')
                continue
            operation, node_name, uid = parts
            uid = int(uid)
            node = self.node_map.get(node_name, None)
            if not node:
                results.append('false')
                continue
            if operation == "1":
                result = self.lock(node, uid)
            elif operation == "2":
                result = self.unlock(node, uid)
            elif operation == "3":
                result = self.upgrade(node, uid)
            else:
                result = False  # Invalid operation
            results.append('true' if result else 'false')
        return results

def main():
    n = int(sys.stdin.readline())
    m = int(sys.stdin.readline())
    q = int(sys.stdin.readline())
    name_list = [sys.stdin.readline().strip() for _ in range(n)]
    queries = [sys.stdin.readline().strip() for _ in range(q)]
    tree = Tree(m, n, name_list)
    results = tree.process_queries(queries)
    for res in results:
        print(res)

if __name__ == "__main__":
    main()

"""
Question 1: Locking the Tree of Space
You have a world map represented as an M-Ary tree (sample tree below):

You need to define three operations on it:
lock(X, uid)
unlock(X, uid)
upgradeLock(X, uid)
Where X is the name of a node in the tree (which is unique), and uid is the user performing the operation.
Definitions for the Operations:
Lock(X, uid):
Lock takes exclusive access on the subtree rooted at X. Once lock(X, uid) succeeds:
lock(A, anyUserId) should fail (returns false), where A is a descendant of X.
lock(B, anyUserId) should fail (returns false), where X is a descendant of B.
Lock operation cannot be performed on a node that is already locked, i.e., lock(X, anyUserId) should fail (returns false).
Unlock(X, uid):
Unlock reverts what was done by the lock operation. It can only be called on the same node on which user uid had called a lock before. Returns true if successful.
UpgradeLock(X, uid):
This helps the user uid upgrade their lock to an ancestor node. It is only possible if the node X already has locked descendants and all of them are locked by the same user uid. Upgrade should fail if there is any node which is a descendant of X that is locked by a different user. A successful upgrade will "lock" the node. UpgradeLock must not violate the consistency model required by Lock/Unlock.
Notes:
The number of nodes in the tree, N, is very large, so optimize the time complexity for the above algorithms.
Input Format:
The first line contains the number of nodes in the tree (N).
The second line contains the number of children per node (m in an m-ary tree).
The third line contains the number of queries (Q).
Next N lines contain the NodeName (string) in the m-Ary Tree.
The next Q lines contain queries in the format: OperationType NodeName UserId.
Operation Type:
1 for Lock
2 for Unlock
3 for UpgradeLock
NodeName: The name of any node (unique) in the m-Ary Tree.
UserId: An integer representing a unique user.
Example Input:
7
2
3
World
Asia
Africa
China
India
South Africa
Egypt
1 China 9
2 India 9
3 Asia 9


This input represents a 2-ary tree with 7 nodes as follows:
           World
           /     \
      Asia        Africa
     /    \       /     \
 China  India  South Africa  Egypt


Additional Notes:
The tree is always fully balanced.
Constraints:
(1 < N < 5 \times 10^5)
(1 < m < 30)
(1 < Q < 5 \times 10^5)
(1 < \text{length of NodeName} < 20)
Optimize the time complexity:
Lock: (O(\log_m N))
Unlock: (O(\log_m N))
UpgradeLock: (O(\text{numberOfLockedNodes} \times \log_m N))
The lock operation on an already locked node should fail.
Once UpgradeLock(X, uid) succeeds on X, it is equivalent to X being locked by uid. As per the definition, Lock(A/B, anyUser) should fail, and Unlock(X, uid) should also work.
Upgrade lock operation on a node having no locked descendants should fail, and upgrade lock on an already locked node should also fail.