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

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;
}
};

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