以下是基于队列的层次遍历实现: fromcollectionsimportdequedeflevel_order_traversal(root):ifnotroot:return[]result=[]queue=deque([root])whilequeue:current_node=queue.popleft()result.append(current_node.value)ifcurrent_node.left:queue.append(current_node.left)ifcurrent_node.right:queue.append(current_no...
LeetCode 0102. Binary Tree Level Order Traversal二叉树的层次遍历【Medium】【Python】【BFS】 Problem LeetCode Given a binary tree, return thelevel ordertraversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree[3,9,20,null,null,15,7], 3 ...
classTreeNode:def__init__(self,value):self.value=value self.left=Noneself.right=Nonedeflevel_order_traversal(root):ifnotroot:return[]queue=[root]result=[]whilequeue:level_size=len(queue)curr_level=[]for_inrange(level_size):node=queue.pop(0)curr_level.append(node.value)ifnode.left:queue...
https://leetcode.com/problems/binary-tree-level-order-traversal/ 题意分析: 宽度优先搜索一颗二叉树,其中同一层的放到同一个list里面。比如: 3 / \ 9 20 / \ 15 7 返回 [ [3], [9,20], [15,7] ] 题目思路: 新定义一个函数,加多一个层数参数,如果目前答案的列表答案个数等于层数,那么当前层数...
(self, root): # 针对N-ary Tree的方法,非常漂亮,前面几个 level_order 都是参考这个 # https://leetcode.com/problems/n-ary-tree-level-order-traversal/description/ q, ret = [root], [] while any(q): ret.append([node.val for node in q]) q = [child for node in q for child in ...
traversal(self) #迭代器 forall(self,op) #op的遍历操作 树的抽象数据类型与二叉树类似,不过在具体实现时,实现策略会有所不同,这里先给出ADT: ADT Tree: Tree(self, data,forest) #树的初始化 is_empty(self) #空树判断 num_nodes(self) #返回结点数据 ...
root.left.left = TreeNode(4) root.left.right = TreeNode(5) result = level_order_traversal(...
节点的层级(Level):该节点的父节点数量+1。 节点的度(Degree):该节点的子节点数量。 二,二叉树的基础概念 每个节点最多有两个子节点的树被称为二叉树。 在给定的二叉树中,任何级别的最大节点数为2l-1,其中"l"是级别编号。 树有很多种,二叉树只是其中一种典型结构,二叉树与其他种类树的区别: ...
Collection: frequent traversal does not care about the order The time complexity of the in operation of lists and tuples is linear, and the in operation of sets and dictionaries is constant.9.充分利用and和or惰性求值特点，减少不必要的计算>python惰性运算特性：>>示例if A and B:print(A)#若A...
defrow_order(tree):#print(tree._data)queue=[]queue.append(tree)whileTrue:ifqueue==[]:breakprint(queue[0]._data)first_tree=queue[0]iffirst_tree._left!=None:queue.append(first_tree._left)iffirst_tree._right!=None:queue.append(first_tree._right)queue.remove(first_tree)if__name__=='...