[LeetCode] Binary Tree Right Side View 二叉树的右侧视图-白红宇

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].

Credits:

Special thanks to for adding this problem and creating all test cases.

这道题要求我们打印出二叉树每一行最右边的一个数字，实际上是求二叉树层序遍历的一种变形，我们只需要保存每一层最右边的数字即可，可以参考我之前的博客，这道题只要在之前那道题上稍加修改即可得到结果，还是需要用到数据结构队列queue，遍历每层的节点时，把下一层的节点都存入到queue中，每当开始新一层节点的遍历之前，先把新一层最后一个节点值存到结果中，代码如下：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector
      
        rightSideView(TreeNode *root) {        vector
       
         res;        if (!root) return res;        queue
        
          q;        q.push(root);        while (!q.empty()) {            res.push_back(q.back()->val);            int size = q.size();            for (int i = 0; i < size; ++i) {                TreeNode *node = q.front();                q.pop();                if (node->left) q.push(node->left);                if (node->right) q.push(node->right);            }        }        return res;    }};

本文转自博客园Grandyang的博客，原文链接：，如需转载请自行联系原博主。