对于图的BFS也是一样滴～ 与Tree的BFS区别如下: tree只有1个root,而图可以有多个源点,所以首先需要把多个源点都入队。 tree是有向的因此不需要标志是否访问过,而对于无向图来说,必须得标志是否访问过! 并且为了防止某个节点多次入队,需要在入队之前就将其设置成已访问! 2、例题1(LeetCode1162题) (1)题目描述...
classSolution{public:vector<int>t;vector<vector<int>>ans;voiddfs(boolchoosePre,intcur,vector<int>&nums){if(cur==nums.size()){ans.push_back(t);return;}dfs(false,cur+1,nums);if(!choosePre&&cur>0&&nums[cur-1]==nums[cur]){return;// 没有选择上一个元素 & 不是第一个元素 & 上一个...
One employee has at most onedirectleader and may have several subordinates. The maximum number of employees won't exceed 2000. 经典BFS:tree的层序遍历思想 """# Employee info class Employee(object): def __init__(self, id, importance, subordinates): # It's the unique id of each node. # ...
1、树的蛇形走位(遍历):Binary Tree Zigzag Level Order Traversal - LeetCode Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree [3,9,20,...
leetcode二叉树-二叉树的最大深度 dfs、bfs,packagebinarytree.maxDepth;importbinarytree.untils.GenerateTreeNode;importbinarytree.untils.TreeNode;importjava.util.ArrayList;importjava.uti
visited 的主要作用是防止走回头路,大部分时候都是必须的。但是像一般的二叉树结构,没有子节点到父节点的指针,不会走回头路就不需要 visited。 二、BFS 解决二叉树的最小深度 给定一个二叉树,找出其最小深度。 最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
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 ...
【刷题】leetcode 102 二叉树层级遍历, Binary Tree Level order Traversal, BFS,python3 370 -- 8:23 App 【刷题】leetcode 1197 马的最短路 knight shortest path, 双向BFS,python3 593 -- 8:40 App 【刷题】leetcode 145 非递归后续遍历 Binary Tree Postorder Traversal,有限状态机,python3 2.7万 180...
look at a specific move, and eliminate it,then backtrack to the next possible move, eliminate it, etc.How to Implement DFS and BFS DFS In tree structure, DFS means we always start from a root node and try to reach the leaf node as direct as possible before we have to backtrack.
// 1.bfspublicList<Integer>rightSideView(TreeNoderoot){List<Integer>res=newArrayList<>();if(root==null)returnres;Queue<TreeNode>queue=newLinkedList<>();queue.add(root);while(!queue.isEmpty()){intn=queue.size();for(inti=0;i<n;i++){TreeNodenode=queue.poll();//记录每层的最后一个元...