[Leetcode] Binary Tree Right Side view two forks

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            <---/   2     3         <---\       5     4       <---

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 traversal of a deformation, we just need to save the rightmost number of each layer can be referred to my previous blog binary tree level order traversal two fork trees sequence traversal, This problem as long as the previous problem can be slightly modified to get results, or need to use the data structure queue, traversing each layer of nodes, the next layer of nodes are deposited into the queue, each time the start of a new layer of node traversal, the first new layer of the last node value is stored in the results, the code is as follows:

/** Definition for binary tree * struct TreeNode {* int val; * TreeNode *left; * TreeNode *right; * T Reenode (int x): Val (x), left (null), right (NULL) {} *}; */classSolution { Public: Vector<int> Rightsideview (TreeNode *root) {Vector<int>Res; if(!root)returnRes; Queue<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 ();                Q.pop (); if(node->left) Q.push (node->Left ); if(node->right) Q.push (node->Right ); }        }        returnRes; }};

[Leetcode] Binary Tree Right Side view two forks