一、問題
利用二叉樹的結構對Huffman樹進行編碼，實現最短編碼
二、解決
1 # 構建節點類
2 class TreeNode: 3 def __init__(self, data): 4 """
5 :data is a tuple the first element is value and the second is priority 6 :param data: 7 """
8 self.value = data[0] 9 self.priority = data[1] 10 self.left_child = None 11 self.right_child = None 12 self.code = ""
13
14
15 # 創建樹節點隊列的函數
16 def create_node_queue(codes): 17 queue = [] 18 for code in codes: 19 queue.append(TreeNode(code)) 20 return queue 21
22
23 # 在隊列中間添加新的節點元素並保證優先度從大到小排列
24 def add_queue(queue, node_new): 25 if len(queue) == 0: 26 return [node_new] 27 for i in range(len(queue)): 28 if queue[i].priority >= node_new.priority: 29 return queue[:i] + [node_new] + queue[i:] 30 return queue + [node_new] 31
32
33 # 節點隊列類
34 class NodeQueue: 35 def __init__(self, code): 36 self.queue = create_node_queue(code) 37 self.size = len(self.queue) 38
39 def add_node(self, node): 40 self.queue = add_queue(self.queue, node) 41 self.size += 1
42
43 def pop_node(self): 44 self.size -= 1
45 return self.queue.pop(0) 46
47
48 # 各個字符在字符串中出現的次數 即計算優先度
49 def frequent_char(string_s): 50 store_d = {} 51 for c in string_s: 52 if c not in store_d: 53 store_d[c] = 1
54 else: 55 store_d[c] += 1
56 return sorted(store_d.items(), key=lambda x: x[1]) 57
58
59 # 創建Huffman樹
60 def create_huffman_tree(node_queue): 61 while node_queue.size != 1: 62 node1 = node_queue.pop_node() 63 node2 = node_queue.pop_node() 64 r_1 = TreeNode([None, node1.priority + node2.priority]) 65 r_1.left_child = node1 66 r_1.right_child = node2 67 node_queue.add_node(r_1) 68 return node_queue.pop_node() 69
70
71 code_dict1 = {} 72 code_dict2 = {} 73
74
75 # 由Huffman樹得到的Huffman編碼表
76 def huffman_code_dict(head, x): 77 # global code_dict, code_list
78 if head: 79 huffman_code_dict(head.left_child, x + "0") 80 head.code += x 81 if head.value: 82 code_dict2[head.code] = head.value 83 code_dict1[head.value] = head.code 84 huffman_code_dict(head.right_child, x + "1") 85
86
87 # 字符串編碼
88 def trans_encode(string_s): 89 # global code_dict1
90 trans_code = ""
91 for c in string_s: 92 trans_code += code_dict1[c] 93 return trans_code 94
95
96 # 字符串解碼
97 def trans_decode(string_s): 98 # global code_dict1
99 code = ""
100 answer = ""
101 for c in string_s: 102 code += c 103 if code in code_dict2: 104 answer += code_dict2[code] 105 code = ""
106 return answer
三、總結
利用Huffman樹的編碼形式可以進行數據的壓縮，因此Huffman的應用也很廣泛。在此記錄一下方便以后查看。