stairs side view

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .Using queues, hierarchy traversal, the last node of each layer./*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode RI Ght * TreeNode (int x) {val = x;} }*/ Public classSo

Title Description:Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered From top to bottom.For example:Given The following binary tree,1 / \2 3 \ \5 4 You should return [1, 3, 4].Ideas:BFS, save the rightmost node.Implementation code:/** * Definition for a binary tree node. * public class TreeNode {* public int val, * public TreeNode left, * public TreeNode right, public Tree Node

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see Ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .1 PublicListRightsideview (TreeNode root) {2listNewArraylist//Save the result to this variable3 if(Root = =NULL)returnRes;4 5linkedlistNewLinkedlist(); 6 Queue.add (root); 7 intcurlevcnt = 1; 8

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .==============The method of using BFS to traverse the binary tree,QueueCurr/next records the number of nodes to traverse in the current layer, and the number of nodes to traverse below====/** Definition for a binary tree n

{ Public: Vectorint> Rightsideview (treenode*root) {Vectorint>ret; TreeNode* current =Root; intdepth =0;//depth so far intDepthgap; StackNodestack; Stackint>Depthstack; while(current) {Ret.push_back ( current-val); Depth++; if(current->Right ) { if(current->Left ) {Nodestack.push ( current-Left ); Depthstack.push (depth); } Current= current->Right ; } Else{ Current= current->Left ; } } while(!Nodestack.empty ()) { Current=

, level +1); -Levelorder (root->right, result, level +1); - } -};2. Iteration:when it reaches the end of the current level, and the push its value into the result vector.Runtime:4ms.1 /**2 * Definition for binary tree3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8 * };9 */Ten classSolution { One Public: Avectorint> Rightsideview (TreeNode *root) { -vectorint>result; - if(!root)returnresult; the

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] ./*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode RI Ght * TreeNode (int x) {val = x;} }*/ Public classSolution { PublicListRightsideview (TreeNode root) {//sequence tr

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .Hierarchical traversal.1 /**2 * Definition for binary tree3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8 * };9 */Ten classSolution { O

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .Credits:Special thanks to @amrsaqr for adding this problem and creating all test cases.This problem requires us to print out a two-fork tree the rightmost number of each line, in fact, is to find a binary tree sequence tra

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .Credits:Special thanks to @amrsaqr for adding this problem and creating all test cases.Algorithm: The transformation of the hierarchical traversal, except that only the rightmost element of each layer is stored./** * Defin

Given A binary tree, imagine yourself standing on the right side of it, return the values of th E nodes can see ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .#include Binary Tree Right Side View

TopicGiven A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .ResolutionTest instructions: Given a binary tree, return the node sequence (top to bottom) seen by the binary tree from the right.Idea: Hierarchical traversal method. When you traverse to the last node of each layer,

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .Idea: BFSclassSolution { Public: Vectorint> Rightsideview (TreeNode *root) {Vectorint>ans; if(Root = NULL)returnans; QueueQ; Q.push (root); while(!Q.empty ()) {Ans.push_back (Q.front ()-val); intpos = Q.siz

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E ordered from top to bottom.For example:Given The following binary tree, 1 You should return [1, 3, 4] .BFS traverses the tree, traversing from right to left, recording only the first one on each level./*** Definition for binary tree* Public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode (int x) {val =

The server side looks at the log shell script (where path1 and path2 are replaced with the path feature name, and the path followed by "Tail-f" is replaced with the log file path for the path feature name):#! /bin/SH#Check log.Echo "which path (G/API):"Read PathEcho "Keyword:"Read keywordif["$path"=='g']; Then Tail-F. /project/g.caipiao.163. Com/log/duobao.log |grep-I."$keyword"elif["$path"=='API']; Then Tail-F. /project/api.g.caipiao.16

" :, "userverify":, "Verifyisopen":, "Version": "", "Vipexpire":, "visits":}See here, you should know the login token, such as sensitive information, DEMOVIP (is a long time ago BEGGARVIP)3. Local_music contains 10 fieldsThis table is a summary of local music and downloaded music.4. play_history contains 13 fields, primarily a summary of the playlist's music information.5. The last table Xm_user is mainly to save the user informationAlthough it is not known what will happen when this information

First, the topic1, examining2. AnalysisA binary tree, looking at him from the right, sees that the first element of each layer is saved up.Second, the answer1, Ideas:Method One,A Queue is used for hierarchical traversal, and the rightmost element of one layer is acquired each time. PublicListRightsideview (TreeNode root) {if(Root = =NULL) return NULL; ListNewArraylist(); QueueNewLinkedlist(); Queue.offer (root); while(!Queue.isempty ()) { intSize =qu

The first one is BFS. Public classSolution { PublicListRightsideview (TreeNode root) {ListNewArraylist(); if(Root = =NULL) { returnresult; } QueueNewLinkedlist(); Queue.offer (root); while(!Queue.isempty ()) { intSize =queue.size (); Result.add (Queue.peek (). val); for(inti = 0; i ) {TreeNode node=Queue.poll (); if(Node.right! =NULL) {queue.offer (node.right); } if(Node.left! =NULL) {queue.offer (node.left); } } }

A simple application of level-order traversal. Just push the last node of each level into the result.The code is as follows.1 classSolution {2 Public:3vectorint> Rightsideview (treenode*root) {4vectorint>Right ;5 if(!root)returnRight ;6Queuetovisit;7 Tovisit.push (root);8 while(!Tovisit.empty ()) {9treenode* Rightnode =Tovisit.back ();TenRight.push_back (Rightnode,val); One intnum =tovisit.size (); A for(inti =0; i ) { -treenode* node =Tovisit.front ();

I saw a lot of BFS based solutions. And my alternative solution is this mirror-ed BST iterator one, with some book-keeping:classSolution { Public: TypeDef pairint>Rec; Vectorint> Rightsideview (TreeNode *root) {Vectorint>v; if(!root)returnv; StackStk; // intMax_d =0; TreeNode*p =Root; while(P) {Stk.push (Rec (p),++max_d)); V.push_back (P-val); P= p->Right ; } // while(!Stk.empty ()) {Rec P0=stk.top (); Stk.pop (); if(p0.first->Left ) {TreeNode*PL = p0.first->Left ;