Binary Tree Right Side View

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].

可用bfs来求解, 关键在于利用标记信息来区分树中的每一层,
然后把每一层最后一个节点的值添加到结果队列中即可.

/**
* 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:
// void rightSideView(const TreeNode* root, vector<int>& res) {
//     if(root == NULL) return ;
//     res.push_back(root->val);
//     if(root->right)
//         rightSideView(root->right, res);
//     else
//         rightSideView(root->left, res);
// }
vector<int> rightSideView(TreeNode* root) {
/// error
// vector<int> res;
// rightSideView(root, res);
// return res;

vector<int> res;
if(root == NULL) return res;
int val;
queue<TreeNode*> q;
// null is a sign variable
TreeNode* null = NULL;
q.push(root);
q.push(null);

// bfs的关键在于找到标记信息(标记每一层)
// bfs中的标记信息为NULL
// 每次遇到NULL, 则更新res
// 值得注意的时, 每次遇到标志信息时,
// 需要检查当前队列是否为空,
// 否则会导致死循环
while(!q.empty()) {
TreeNode* r = q.front();
q.pop();
if(r == null) {
// if q is empty, stop pushing null
if(!q.empty()) {
q.push(null);
}
res.push_back(val);
} else {
val = r->val;
if(r->left) q.push(r->left);
if(r->right) q.push(r->right);
}
}
return res;
}
};