[Leetcode] Binary Tree Right Side View

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:3vector<int> Rightsideview (treenode*root) {4vector<int>Right ;5         if(!root)returnRight ;6Queue<treenode*>tovisit;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 < num; i++) { -treenode* node =Tovisit.front (); - Tovisit.pop (); the                 if(node-to-left) Tovisit.push (node-Left ); -                 if(node-right) Tovisit.push (node-Right ); -             } -         } +         returnRight ; -     } +};

Well, the above code is of BFS. This problem can still is solved using DFS. The code is as follows. Play with it-see how it works:-)

1 classSolution {2  Public:3vector<int> Rightsideview (treenode*root) {4vector<int>Right ;5         if(!root)returnRight ;6Rightview (Root, right,0);7     }8 Private:9     voidRightview (treenode* node, vector<int>& Right,intLevel ) {Ten         if(!node)return; One         if(Level = =right.size ()) ARight.push_back (Node-val); -Rightview (node-right, right, level +1); -Rightview (node-I, right, level +1); the     } -};

[Leetcode] Binary Tree Right Side View