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 @amrsaqr 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 Binary Tree Level Order Traversal 二叉树层序遍历，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<int> rightSideView(TreeNode *root) { vector<int> res; if (!root) return res; queue<TreeNode*> 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：二叉树的右侧视图[LeetCode] Binary Tree Right Side View ，如需转载请自行联系原博主。
