fromcollectionsimportdeque# 定义二叉树节点classTreeNode:def__init__(self,val=0,left=None,right=None):self.val=valself.left=leftself.right=rightdeflevelOrder(root):# 如果根节点为空,返回空列表ifnotroot:return[]# 初始化队列和结果列表queue=deque([root])result=[]whilequeue:level_size=len(queue...
使用递归,根据树深度判断节点值应该添加入集合的哪个位置。 publicList<List<Integer>>levelOrder(TreeNode root){ List<List<Integer>> res =newArrayList<>();if(root ==null)returnres; travel(root,res,0);returnres; }privatevoidtravel(TreeNode root,List<List<Integer>> res,intdepth){if(root ==null...
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 ...
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7], 代码语言:javascript 代码运行次数:0 运行 AI代码解释 3/\920/\157 return its level order traversal as: 代码...
push(NULL); 29 int nLevelCount = 1; 30 while (true) { 31 TreeNode *pTemp = tree_queue.front(); 32 tree_queue.pop(); 33 if (pTemp == NULL) { 34 if (nLevelCount%2 == 0) { //if the num of level is odd, swap the ivec; 35 Swap(ivec); 36 } 37 tree_vector.push_...
return its bottom-up level order traversal as: [ [15,7], [9,20], [3] ] 1. 2. 3. 4. 5. # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None ...
return its bottom-up level order traversal as: [ [15,7], [9,20], [3] ] 1. 2. 3. 4. 5. 分析: 这道题目跟上道题目很相似 Leetcode: Binary Tree Level Order Traversal ,唯一不同的就是返回结果是从子叶节点到根节点,所以我们只需要将结果翻转下就好了!
Can you solve this real interview question? Binary Tree Level Order Traversal - Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). Example 1: [https://assets.leetcode.c
Binary Tree Inorder Traversal Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. 栈迭代 复杂度 时间O(b^(h+1)-1) 空间 O(h) 递归栈空间 对于二叉树b=2 ...
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/ 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。解法一：层序遍历首先，如果根节点为空，直接返回空的结果集。如果根节点不为空，通过队列来遍历每一层的节点，具体处理过程如下：...