Binary Tree Right Side View--leetcode

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] .

Idea: the equivalent of a hierarchical traversal, just note that the last element of the output layer.

void Rightsideview (Bintree *root) {vector<int> VEC;     Deque<bintree*> de;     if (root = NULL) return;     int curnum = 1;     int nexnum=0;     De.push_back (root);     bintree* temp;             while (!de.empty ()) {while (Curnum >0) {temp = De.front ();             if (Curnum ==1) vec.push_back (Temp->value);             De.pop_front ();             curnum--;                if (temp->left! = NULL) {de.push_back (temp->left);                        nexnum++;                 } if (temp->right! = NULL) {de.push_back (temp->right);             nexnum++;         }} curnum = Nexnum;                  nexnum=0;     } for (int i=0;i<vec.size (); i++) cout<<vec[i]<< "";  cout<<endl; }

Binary Tree Right Side View--leetcode