[Leetcode]199.binary Tree Right Side View

Topic

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

Hierarchical traversal

Code

    /*-------------------------------------------------------------------* Date: 2014-04-05 * SJF0115 * Title: 1 99.Binary Tree Right Side View * Source: https://leetcode.com/problems/binary-tree-right-side-view/* Result: AC * source : Leetcode * Summary:--------------------------------------------------------------------*/    #include <iostream>    #include <vector>    #include <queue>    using namespace STD;structtreenode{intVal        TreeNode *left;        TreeNode *right; TreeNode (intx): Val (x), left (null), right (null) {}};classSolution { Public: vector<int>Rightsideview (TreeNode *root) { vector<int>Resultif(Root = =nullptr){returnResult }//if             queue<TreeNode*>Cur queue<TreeNode*>Next            Cur.push (root);            Result.push_back (Root->val); TreeNode *node,*pre;//Hierarchical Traversal             while(!cur.empty ()) {pre =nullptr; while(!cur.empty ())                    {node = Cur.front (); Cur.pop ();if(Node->left)                        {Next.push (node->left);                    Pre = node->left; }//if                    if(Node->right)                        {Next.push (node->right);                    Pre = node->right; }//if}//while                if(Pre! =nullptr) {Result.push_back (pre->val); }//ifSwap (Cur,next); }//while            returnResult }    };intMain () {solution solution; TreeNode *root =NewTreeNode (1); Root->left =NewTreeNode (2); Root->right =NewTreeNode (3); Root->left->right =NewTreeNode (5); Root->right->right =NewTreeNode (4); vector<int>result = Solution.rightsideview (root); for(inti =0; i < result.size (); ++i) {cout<<result[i]<<" "; }//for    cout<<endl;return 0;}

Run time

[Leetcode]199.binary Tree Right Side View