练习内容：
判断一个二叉树T1是否是另一颗二叉树T2的子树。(T2>T1)
要求：时间复杂度O(N)，空间复杂度O(1)
正文内容提要：
1.morris遍历实现T1，T2的先序序列化。
2.kmp算法判断T1与T2的序列化结果(T2>T1)
（1）如果seralize(T1)是seralize(T2)的子串，则T1是T2的子树
（2）否则，T1不是T2的子树
1.创建类描述树结构
1 class MyProperty: 2 def __init__(self, get_attrib_func): 3 self.get_attrib_func = get_attrib_func 4 self.set_attrib_func = None 5 6 def __set__(self, instance, value): 7 if not self.set_attrib_func: 8 raise AttributeError("can't set attribute") 9 self.set_attrib_func(instance, value) 10 11 def __get__(self, instance, owner): 12 return self.get_attrib_func(instance) 13 14 def setter(self, set_attrib_func): 15 self.set_attrib_func = set_attrib_func 16 return self 17 18 19 class Node: 20 def __init__(self, value): 21 self.value = value 22 self.__left = None 23 self.__right = None 24 25 @MyProperty 26 def left(self): 27 return self.__left 28 29 @left.setter 30 def left(self, node): 31 self.__left = node 32 33 @MyProperty 34 def right(self): 35 return self.__right 36 37 @right.setter 38 def right(self, node): 39 self.__right = node
2.使用morris遍历，实现二叉树的先序序列化
1 def morris_pre_serialize(head): 2 if not head: 3 return 4 string = [] 5 cur = head 6 most_right = None 7 while cur != None: 8 most_right = cur.left 9 if most_right != None: 10 while most_right.right != None and most_right.right != cur: 11 most_right = most_right.right 12 13 if most_right.right == None: # 第一次到达cur节点 14 string.append("{}{}".format(cur.value, "_")) 15 most_right.right = cur 16 cur = cur.left 17 continue 18 else: 19 most_right.right = None # 第二次到达cur节点 20 string.append("#_") 21 22 else: 23 string.append(str(cur.value) + "_") 24 string.append("#_") 25 cur = cur.right 26 else: 27 string.append("#_") 28 29 return "".join(string)
3.实现kmp算法
1 def kmp(str_x, str_y): 2 3 def __get_nexts(str_x): 4 if str_x is None: 5 return None 6 7 length = len(str_x) 8 if length == 1: 9 return [-1] 10 11 nexts = [0] * length 12 nexts[0] = -1 13 nexts[1] = 0 14 i1, i2 = 2, 1 15 while i1 < length: 16 if str_x[i1 - 1] == str_x[nexts[i2]]: 17 nexts[i1] = nexts[i2] + 1 18 i2, i1 = i1, i1 + 1 19 elif nexts[i2] != -1: 20 i2 = nexts[i2] 21 else: 22 nexts[i1] = 0 23 i2, i1 = i1, i1 + 1 24 return nexts 25 26 27 if str_x is None or str_y is None or len(str_y) < 1 or len(str_x) < len(str_y): 28 return -1 29 30 len_str_x, len_str_y = len(str_x), len(str_y) 31 32 i1, i2 = 0, 0 33 34 nexts = __get_nexts(str_y) 35 36 while i1 < len_str_x and i2 < len_str_y: 37 if str_x[i1] == str_y[i2]: 38 i1, i2 = i1 + 1, i2 + 1 39 elif nexts[i2] != -1: 40 i2 = nexts[i2] 41 else: 42 i1 += 1 43 return i1 - i2 if i2 == len_str_y else -1
4.结合morris遍历，kmp子串判断，实现子树的判断。时间复杂O(N), 空间复杂度O(1)
1 def is_sub_tree(head1, head2): 2 if head1 is None or head2 is None: 3 return False 4 5 str1 = morris_pre_serialize(head1) 6 str2 = morris_pre_serialize(head2) 7 8 str1, str2 = (str2, str1) if len(str1) < len(str2) else (str1, str2) 9 10 ret = kmp(str1, str2) 11 12 return True if ret != -1 else False
5.简单测试代码
1 if __name__ == "__main__": 2 tree_1 = Node(1) 3 tree_1.left = Node(2) 4 tree_1.right = Node(3) 5 tree_1.left.left = Node(4) 6 tree_1.left.right = Node(5) 7 tree_1.right.left = Node(6) 8 tree_1.right.right = Node(7) 9 10 tree_2 = Node(3) 11 tree_2.left = Node(6) 12 tree_2.right = Node(7) 13 tree_2.left.right = Node(8) 14 # tree_2 is not tree_1's sub tree 15 print(is_sub_tree(tree_2, tree_1)) # False 16 print("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~") 17 tree_3 = Node(3) 18 tree_3.left = Node(6) 19 tree_3.right = Node(7) 20 # tree_3 is tree_1's sub tree 21 print(is_sub_tree(tree_1, tree_3)) # True