生成二叉樹

# 節點類 class Node(object):def __init__(self, x):self.val = xself.left = Noneself.right = None # 樹生成代碼 def generate_tree(vals):if len(vals) == 0:return Noneque = [] # 定義隊列fill_left = True # 由于無法通過是否為 None 來判斷該節點的左兒子是否可以填充，用一個記號判斷是否需要填充左節點for val in vals:node = Node(val) if val else None # 非空值返回節點類，否則返回 Noneif len(que)==0:root = node # 隊列為空的話，用 root 記錄根結點，用來返回que.append(node)elif fill_left:que[0].left = nodefill_left = False # 填充過左兒子后，改變記號狀態if node: # 非 None 值才進入隊列que.append(node)else:que[0].right = nodeif node:que.append(node)que.pop(0) # 填充完右兒子，彈出節點fill_left = True # return root # 定義一個dfs打印中序遍歷 def dfs(node):if node is not None:dfs(node.left)print(node.val, end=' ')dfs(node.right) # 定義一個bfs打印層序遍歷 def bfs(node):que = []que.append(node)while que:l = len(que)for _ in range(l):tmp = que.pop(0)print(tmp.val, end=' ')if tmp.left:que.append(tmp.left)if tmp.right:que.append(tmp.right)print('|', end=' ')# test null = None vals = [3,9,20,null,null,15,7] tree = generate_tree(vals) print('中序遍歷:') dfs(tree) # 9 3 15 20 7 print('\n層序遍歷:') bfs(tree) # 3 | 9 20 | 15 7 |

生成列表

#定義節點 class ListNode():def __init__(self, x):self.val = xself.next = None #將傳入的數組轉化為鏈表 def create_linked_list(arr):head = ListNode(arr[0])cur = headfor i in range(1, len(arr)):cur.next = ListNode(arr[i])cur = cur.nextreturn head #傳入鏈表頭節點，以數組形式返回 def print_linked_list(head):cur = headres = []while cur:res.append(cur.val)cur = cur.nextreturn res class Solution():def mergeTwoLists(self, l1, l2):pre = ListNode(0)head = prewhile l1 and l2:if l1.val >= l2.val:pre.next = l2l2 = l2.nextelse:pre.next = l1l1 = l1.nextpre = pre.nextpre.next = l1 if l1 else l2return head.next if __name__ == "__main__":head1 = create_linked_list([1, 2, 4])head2 = create_linked_list([1, 3, 4])solution = Solution()sorted_lists = solution.mergeTwoLists(head1, head2)print(print_linked_list(sorted_lists)) #輸出：[1, 1, 2, 3, 4, 4]

注：放到本地可以直接跑哦~

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