dom ×2 python ×2 stl ×2 c ×1 depth-first-search ×1 dictionary ×1 directory-traversal ×1 directory-walk ×1 glob ×1 graph ×1 iterator ×1 language-agnostic ×1 list ×1 matrix ×1 next ×1 os.walk ×1 reference ×1 symlink ×1 symlink-traversal ×1 tree ×1 xml ×1«...
# print(type(root), type(curr_level)) # (<class 'precompiled.treenode.TreeNode'>, <type 'list'>) # print(curr_level) # 作为list,却并不能遍历整个树 while curr_level: level_list = [] next_level = [] for temp in curr_level: level_list.append(temp.val) if temp.left: next_leve...
objsizecan calculate the size of an object’s entire subtree in bytes regardless of the type of objects in it, and its depth. Here is a complex data structure, for example, that include a self reference: my_data=list(range(3)),list(range(3,6))classMyClass:def__init__(self,x,y)...
Python3代码 # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def levelOrder(self, root: TreeNode) -> List[List[int]]: # solution two: 递归 res = [] def helper(root, ...
The total number of nodes is at most 5000. 题目大意 N叉树的层次遍历。 解题方法 方法一:BFS 首先得明白,这个N叉树是什么样的数据结构定义的。val是节点的值,children是一个列表,这个列表保存了其所有节点。 层次遍历比较好理解,就是每层的值保存在一个list中,总的再返回一个list即可。
AC代码(Python) 1#Definition for a binary tree node2#class TreeNode:3#def __init__(self, x):4#self.val = x5#self.left = None6#self.right = None78classSolution:9#@param root, a tree node10#@return a list of integers11defiterative_inorder(self, root, list):12stack =[]13whilero...
def postorderTraversal(self, root: TreeNode) -> List[int]: res: List[int] = [] dummyRoot = TreeNode(-1, root) cur = dummyRoot while cur: # here is the preorder slot if cur.left: # find the inorder predecessor of cur
python class Solution: """ @param matrix: An array of integers @return: An array of integers """ def printZMatrix(self, matrix): if not matrix or not matrix[0]: return [] m = len(matrix) n = len(matrix[0]) to_upper_right = True res = [] i = 0 j = 0 for _ in range...
:rtype: List[List[int]] """ ifnotroot: return[] ans=[[root.val]] queue=deque([root]) whilequeue: levelans=[] for_inxrange(0,len(queue)): root=queue.popleft() ifroot.left: levelans.append(root.left.val) queue.append(root.left) ...
Linear data structures like arrays, stacks, queues, and linked list have only one way to read the data. But a hierarchical data structure like a tree can be traversed in different ways. Tree traversal Let's think about how we can read the elements of the tree in the image shown above...