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70 lines
1.8 KiB
Markdown
70 lines
1.8 KiB
Markdown
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# LeetCode 第 199 号问题：二叉树的右视图
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> 本文首发于公众号「五分钟学算法」，是[图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>)系列文章之一。
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>
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> 个人网站：[https://www.cxyxiaowu.com](https://www.cxyxiaowu.com)
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题目来源于 LeetCode 上第 199 号问题：二叉树的右视图。题目难度为 Medium，目前通过率为 57.5% 。
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### 题目描述
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给定一棵二叉树，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。
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**示例:**
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```
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输入: [1,2,3,null,5,null,4]
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输出: [1, 3, 4]
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解释:
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1 <---
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/ \
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2 3 <---
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\ \
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5 4 <---
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```
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### 题目解析
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与之前[二叉树的层次遍历](https://xiaozhuanlan.com/topic/8579460312)类似的，该问题需要用到**队列**,
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- 建立一个 queue
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- 遍历每层的节点时，把下一层的节点都存入到 queue 中
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- 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
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### 动画描述
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### 代码实现
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```
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class Solution {
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public:
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vector<int> rightSideView(TreeNode *root) {
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vector<int> res;
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if (!root) return res;
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queue<TreeNode*> q;
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q.push(root);
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while (!q.empty()) {
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res.push_back(q.back()->val);
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int size = q.size();
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for (int i = 0; i < size; ++i) {
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TreeNode *node = q.front();
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q.pop();
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if (node->left) q.push(node->left);
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if (node->right) q.push(node->right);
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}
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}
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return res;
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}
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};
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```
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