https://www.cnblogs.com/lz87/p/10159724.html
X. https://www.geeksforgeeks.org/locking-and-unlocking-of-resources-in-the-form-of-n-ary-tree/
The idea is to augment tree with following three fields:
BinaryTreeLockTemplate.java
Solution 1.
is_locked(): O(1)
This problem was asked by Google.
Implement locking in a binary tree. A binary tree node can be locked or unlocked only if all of its descendants or ancestors are not locked.
Design a binary tree node class with the following methods:
is_locked, which returns whether the node is lockedlock, which attempts to lock the node. If it cannot be locked, then it should return false. Otherwise, it should lock it and return true.unlock, which unlocks the node. If it cannot be unlocked, then it should return false. Otherwise, it should unlock it and return true.
You may augment the node to add parent pointers or any other property you would like. You may assume the class is used in a single-threaded program, so there is no need for actual locks or mutexes. Each method should run in O(h), where h is the height of the tree.
Solution 1.
is_locked(): O(1)
lock(): O(m + h)
unlock(): O(m + h)
lock and unlock will take O(m + h) time where m is the number of nodes in the node's subtree (since we have to traverse through all its descendants) and h is the height of the node (since we have to traverse through the node's ancestors).
This problem was asked by Google.
Implement locking in a binary tree. A binary tree node can be locked or unlocked only if all of its descendants or ancestors are not locked.
Design a binary tree node class with the following methods:
is_locked, which returns whether the node is lockedlock, which attempts to lock the node. If it cannot be locked, then it should return false. Otherwise, it should lock it and return true.unlock, which unlocks the node. If it cannot be unlocked, then it should return false. Otherwise, it should unlock it and return true.
You may augment the node to add parent pointers or any other property you would like. You may assume the class is used in a single-threaded program, so there is no need for actual locks or mutexes. Each method should run in O(h), where h is the height of the tree.
X. https://www.geeksforgeeks.org/locking-and-unlocking-of-resources-in-the-form-of-n-ary-tree/
The idea is to augment tree with following three fields:
- A boolean field isLocked (same as above method).
- Parent-Pointer to access all ancestors in O(Log n) time.
- Count-of-locked-descendants. Note that a node can be ancestor of many descendants. We can check if any of the descendants is locked using this count.
Let us see how operations work using this Approach.
- isLock(): Check isLocked of the given node.
- Lock(): Traverse all ancestors. If none of the ancestors is locked, Count-of-locked-descendants is 0 and isLocked is false, set isLocked of current node as true. And increment Count-of-locked-descendants for all ancestors. Time complexity is O(Log N) as there can be at most O(Log N) ancestors.
- unLock(): Traverse all ancestors and decrease Count-of-locked-descendants for all ancestors. Set isLocked of current node as false. Time complexity is O(Log N)
- For each node keep track of the number of nodes in its subtree that are locked
- To lock,
- check child locked count
- check all nodes in the path to parent
- if none of the above are locked, then mark node as locked
- To unlock
- mark node as unlocked
- update all nodes in the path to parent and decrement their lock count
Solution 2 with O(h) lock and unlock
Since solution 1 does not meet the O(h) requirement, we can improve the performance of lock and unlock by adding another field to the node that keeps tracks of the count of locked descendants. That way, we can immediately see whether any of its descendants are locked. This will reduce our
lock and unlock functions to only O(h). We can maintain this field by doing the following:- When locking, if the locking succeeds, traverse the node's ancestors and increment each one's count
- When unlocking, traverse the node's ancestors and decrement each one's count
1 public class BinaryTreeWithLock { 2 private class BinaryTreeNode { 3 int val; 4 boolean locked = false; 5 BinaryTreeNode parent; 6 BinaryTreeNode leftChild; 7 BinaryTreeNode rightChild; 8 int lockedDescendantCount = 0; 9 } 10 11 public boolean is_locked(BinaryTreeNode node) { 12 return node.locked; 13 } 14 public boolean lock(BinaryTreeNode node) { 15 if(is_locked(node)) { 16 return true; 17 } 18 if(!canLockOrUnlock(node)) { 19 return false; 20 } 21 node.locked = true; 22 BinaryTreeNode parentNode = node.parent; 23 while(parentNode != null) { 24 parentNode.lockedDescendantCount += 1; 25 parentNode = parentNode.parent; 26 } 27 return true; 28 } 29 public boolean unlock(BinaryTreeNode node) { 30 if(!is_locked(node)) { 31 return true; 32 } 33 if(!canLockOrUnlock(node)) { 34 return false; 35 } 36 node.locked = false; 37 BinaryTreeNode parentNode = node.parent; 38 while(parentNode != null) { 39 parentNode.lockedDescendantCount -= 1; 40 parentNode = parentNode.parent; 41 } 42 return true; 43 } 44 private boolean canLockOrUnlock(BinaryTreeNode node) { 45 if(node.lockedDescendantCount > 0) { 46 return false; 47 } 48 BinaryTreeNode parentNode = node.parent; 49 while(parentNode != null) { 50 if(parentNode.locked) { 51 return false; 52 } 53 parentNode = parentNode.parent; 54 } 55 return true; 56 } 57 }
public class BinaryTreeLockTemplate {
public static class BinaryTree<T> {
private BinaryTree<T> left, right, parent;
private boolean locked;
private int numChildrenLocks;
public boolean isLock() {
return locked;
}
public void lock() {
if (numChildrenLocks == 0 && !locked) {
// Make sure all parents do not lock.
BinaryTree<T> n = parent;
while (n != null) {
if (n.locked) {
return;
}
n = n.parent;
}
// Lock itself and update its parents.
locked = true;
n = parent;
while (n != null) {
++n.numChildrenLocks;
n = n.parent;
}
}
}
public void unlock() {
if (locked) {
// Unlock itself and update its parents.
locked = false;
BinaryTree<T> n = parent;
while (n != null) {
--n.numChildrenLocks;
n = n.parent;
}
}
}
}
}
A way to implement this would be to augment each node with an
is_locked attribute as well as a parent pointer. We can then implement the methods this way:is_lockedsimply returns the node’s attributelocksearches the node’s children and parents for a trueis_lockedattribute. If it is set to true on any of them, then return false. Otherwise, set the current node’sis_lockedto true and return true.unlocksimply changes the node’s attribute to false. If we want to be safe, then we should search the node’s children and parents as inlockto make sure we can actually unlock the node, but that shouldn’t ever happen.
While
is_locked is O(1) time, lock and unlock will take O(m + h) time where m is the number of nodes in the node’s subtree (since we have to traverse through all its descendants) and h is the height of the node (since we have to traverse through the node’s ancestors).
We can improve the performance of
lock and unlock by adding another field to the node that keeps tracks of the count of locked descendants. That way, we can immediately see whether any of its descendants are locked. This will reduce our lock and unlock functions to only O(h). We can maintain this field by doing the following:- When locking, if the locking succeeds, traverse the node’s ancestors and increment each one’s count
- When unlocking, traverse the node’s ancestors and decrement each one’s count
class LockingBinaryTreeNode(object): def __init__(self, val, left=None, right=None, parent=None): self.val = val self.left = left self.right = right self.parent = parent self._is_locked = False self.locked_descendants_count = 0 def _can_lock_or_unlock(self): if self.locked_descendants_count > 0: return False cur = self.parent while cur: if cur._is_locked: return False cur = cur.parent return True def is_locked(self): return self._is_locked def lock(self): if self._can_lock_or_unlock(): # Not locked, so update is_locked and increment count in all ancestors self._is_locked = True cur = self.parent while cur: cur.locked_descendants_count += 1 cur = cur.parent return True else: return False def unlock(self): if self._can_lock_or_unlock(): self._is_locked = False # Update count in all ancestors cur = self.parent while cur: cur.locked_descendants_count -= 1 cur = cur.parent return True else: return False
Solution 1.
is_locked(): O(1)
lock(): O(m + h)
unlock(): O(m + h)
lock and unlock will take O(m + h) time where m is the number of nodes in the node's subtree (since we have to traverse through all its descendants) and h is the height of the node (since we have to traverse through the node's ancestors).1 public class BinaryTreeWithLock { 2 private class BinaryTreeNode { 3 boolean locked; 4 BinaryTreeNode parent; 5 BinaryTreeNode leftChild; 6 BinaryTreeNode rightChild; 7 } 8 9 public boolean is_locked(BinaryTreeNode node) { 10 return node.locked; 11 } 12 public boolean lock(BinaryTreeNode node) { 13 if(!checkLockPrecondition(node)) { 14 return false; 15 } 16 node.locked = true; 17 return true; 18 } 19 public boolean unlock(BinaryTreeNode node) { 20 if(!checkLockPrecondition(node)) { 21 return false; 22 } 23 node.locked = false; 24 return true; 25 } 26 private boolean checkLockPrecondition(BinaryTreeNode node) { 27 //check descendants 28 boolean left = checkChildLockPrecondition(node.leftChild); 29 if(!left) { 30 return false; 31 } 32 boolean right = checkChildLockPrecondition(node.rightChild); 33 if(!right) { 34 return false; 35 } 36 //check ancestors 37 BinaryTreeNode parentNode = node.parent; 38 while(parentNode != null) { 39 if(parentNode.locked) { 40 return false; 41 } parentNode = parentNode.parent; 42 } 43 return true; 44 } 45 private boolean checkChildLockPrecondition(BinaryTreeNode node) { 46 if(node == null) { 47 return true; 48 } 49 if(node.locked) { 50 return false; 51 } 52 boolean left = checkChildLockPrecondition(node.leftChild); 53 if(!left) { 54 return false; 55 } 56 boolean right = checkChildLockPrecondition(node.rightChild); 57 return right; 58 } 59 }
