原题 层序遍历,从自底向上按层输出. 左→右→中 解法一 : DFS,求出自顶向下的,最后返回时反转一下. class Solution { public: vector<vector<int>> res; vector<vector<int>> levelOrderBottom(TreeNode *root) { int level = 0; dfs(root, level); reverse(res.begin(), res.end()); return…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order t…

翻译 给定一个二叉树,返回从下往上遍历经过的每一个节点的值. 从左往右,从叶子到节点. 比如: 给定的二叉树是 {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 返回它从下往上的遍历结果: [ [15,7], [9,20], [3] ] 原文 Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, lev…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

题目描述: Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

相似题目: 102 103 107 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> levelOrd…

题意是倒过来层次遍历二叉树 下面我介绍下BFS的基本框架,所有的BFS都是这样写的 struct Nodetype { int d;//层数即遍历深度 KeyType m;//相应的节点值 } queue<Nodetype> q; q.push(firstnode); while(!q.empty()){ Nodetype now = q.front(); q.pop(); ........ for(遍历所有now点的相邻点next){ if(!visit[next]) { 访问每个没有访问过…

原文题目: 102. Binary Tree Level Order Traversal 107. Binary Tree Level Order Traversal II 读题: 102. 层序遍历二叉树,每一层作为一个数组,从上到下输出 107.层序遍历二叉树,每一层作为一个数组,反过来从下到上输出 两者只有最后一行的存储方式不一致 class Solution(object): def levelOrder(self, root): """ :type root: Tr…

Binary Tree Level Order Traversal II Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ Given a binary tree, return the bottom-up level order traversal of its nodes' values.…

题目来源 https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). 题意分析 Input: binary tree Output:…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 方法一:DFS 方法二:迭代 日期 [LeetCode] 题目地址:https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ Total Accepted: 55876 Total Submissions: 177210 Difficulty: Easy 题目描述 Given…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order tr…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order tr…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

题目: Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level orde…

从上往下打印二叉树这个是不分行的,用一个队列就可以实现 class Solution { public: vector<int> PrintFromTopToBottom(TreeNode* root) { vector<int> result; queue<TreeNode* > container; if(root == NULL) return result; container.push(root); ){ TreeNode* rot = container.f…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order t…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order t…

这次相对来讲复杂点,题目如下: Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-u…

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level or…

#-*- coding: UTF-8 -*- # Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def levelOrderBottom(self, root):   …

按层输出二叉树,广度优先. 3 / \ 9 20 / \ 15 7 [ [15,7], [9,20], [3] ] /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ //广度优先遍历 class Solut…

给定一个二叉树,返回其节点值自底向上的层次遍历. (即按从叶节点所在层到根节点所在的层,逐层从左向右遍历)例如:给定二叉树 [3,9,20,null,null,15,7],    3   / \  9  20    /  \   15   7返回其自自底向上的层次遍历为:[  [15,7],  [9,20],  [3]]详见:https://leetcode.com/problems/binary-tree-level-order-traversal-ii/description/ Java实现…

val = $value; } * } */ class Solution { private $vals = []; /** * @param TreeNode $root * @return Integer[][] */ function levelOrderBottom($root) { $this->preOrder($root, 0); return array_reverse($this->vals); } function preOrder($node, $level){ if(…