LeetCode好久没有总结过了，今天来看一道二叉树Medium题目

  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]. 

这道题目并不难，找出层次遍历（BFS）每一层最后一个节点，有趣的地方在于如何优雅准确的一次通过。

这里用两个队列，并且使用翻转技巧

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int > img;
        queue<TreeNode *> q[2];
        bool qi=false;
        if(root){
            q[qi].push(root);
        }
        while(true){
            if(q[qi].size()==0){
                break;
            }
            TreeNode *t;
            while(q[qi].size() > 0){
                t = q[qi].front();
                q[qi].pop();
                if(t->left){
                    q[!qi].push(t->left);
                }
                if(t->right){
                    q[!qi].push(t->right);
                }
            }
            img.push_back(t->val);
            qi=!qi;
        }
        return img;
    }
};

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