python多线程死锁现象及解决方法

from threading import Thread,Lock
import time
mutexA=Lock()
mutexB=Lock()
class MyThread(Thread):
    def run(self):
        self.func1()
        self.func2()
    def func1(self):
        mutexA.acquire()
        print('\033[41m%s 拿到A锁\033[0m' %self.name)
        mutexB.acquire()
        print('\033[42m%s 拿到B锁\033[0m' %self.name)
        mutexB.release()
        mutexA.release()
    def func2(self):
        mutexB.acquire()
        print('\033[43m%s 拿到B锁\033[0m' %self.name)
        time.sleep(2)
        mutexA.acquire()
        print('\033[44m%s 拿到A锁\033[0m' %self.name)
        mutexA.release()
        mutexB.release()
if __name__ == '__main__':
    for i in range(10):
        t=MyThread()
        t.start()

Thread-1 拿到A锁
Thread-1 拿到B锁
Thread-1 拿到B锁
Thread-2 拿到A锁 #出现死锁，整个程序阻塞住

from threading import Thread,RLock
import time
mutexA=mutexB=RLock() #一个线程拿到锁，counter加1,该线程内又碰到加锁的情况，则counter继续加1，这期间所有其他线程都只能等待，等待该线程释放所有锁，即counter递减到0为止
class MyThread(Thread):
    def run(self):
        self.func1()
        self.func2()
    def func1(self):
        mutexA.acquire()
        print('\033[41m%s 拿到A锁\033[0m' %self.name)
        mutexB.acquire()
        print('\033[42m%s 拿到B锁\033[0m' %self.name)
        mutexB.release()
        mutexA.release()
    def func2(self):
        mutexB.acquire()
        print('\033[43m%s 拿到B锁\033[0m' %self.name)
        time.sleep(2)
        mutexA.acquire()
        print('\033[44m%s 拿到A锁\033[0m' %self.name)
        mutexA.release()
        mutexB.release()
if __name__ == '__main__':
    for i in range(10):
        t=MyThread()
        t.start()