Java two fork tree traversal right view-leetcode199

The topics are as follows:

The example given in the topic is not very good, it is easy to make people misunderstand the right node to continue to access the good, but the title is not so.

Change to popular meaning: traverse the binary tree by layer and output the most right end node of each layer.

This is understood when a binary tree sequence traversal problem, with a queue to deal with, but the question is how to identify the most right side of each layer node, I think for a long time, the last way to figure out is to use a marker, such as the above example:

Q Team column, F for the tag node, right for the record's best-left node.

Q:1 Flag right:{}

Q:flag 2 3 encounters a marker bit so move the marker bit and save the team header popup data as follows

Q:2 3 Flag Right:{1}

Q:3 Flag 5 Right:{1}

Q:flag 5 4 encounters a marker bit so move the marker bit and save the team header popup data as follows

Q:5 4 flag Right:{1 3}

Q:4 Flag Right:{1 3}

Q:flag encounters a marker bit so move the marker bit and save the data from the team header as follows

Q:flag right:{1 3 4}

At this point, the queue element is found only 1, exit loop return results

The code is as follows:

     Public classTreeNode {intVal;        TreeNode left;        TreeNode right; TreeNode (intx) {val=x; }    }     PublicList<integer>Rightsideview (TreeNode root) {List<Integer> right =NewArraylist<integer>(); if(Root = =NULL)            returnRight ; Queue<TreeNode> q =NewLinkedlist<treenode>(); TreeNode P=Root; TreeNode Flag=NewTreeNode (-99999999);        Q.add (P);        Q.add (flag);  while(Q.size ()! = 1) {p=Q.poll (); if(P.left! =NULL) Q.add (p.left); if(P.right! =NULL) Q.add (p.right); if(Q.peek (). val = =-99999999) {right.add (p.val);                Q.poll ();            Q.add (flag); }        }        returnRight ; }

Here I mark bit began to use-1, later depressed the discovery test concentrated node element has-1, changed to now this, passed.

In addition to the online review of other people's solution, there will be a layer of code all access, and then go to the next layer of elements to find the right end of each layer node, the code is as follows:

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: Avector<int> Rightsideview (TreeNode *root) { -vector<int>Res; -         if(!root)returnRes; theQueue<treenode*>Q; - Q.push (root); -          while(!Q.empty ()) { -Res.push_back (Q.back ()val); +             intSize =q.size (); -              for(inti =0; i < size; ++i) { +TreeNode *node =Q.front (); A Q.pop (); at                 if(node->left) Q.push (node->Left ); -                 if(node->right) Q.push (node->Right ); -             } -         } -         returnRes; -     } in};

C + + written, both ways can be

Java two fork tree traversal right view-leetcode199