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].
求每一层的最右的一个节点，层次遍历树，每一层的最后一个节点就是结果元素
使用一个桶封装树节点，加上层次，当遍历层次更新时，上一个节点就是上一个层次的最后一个节点
typedef struct bucket{
TreeNode *node;
int level;
bucket(TreeNode *n, int l):node(n), level(l){}
} bucket, *Bucket;
class Solution {
public:
vector<int> rightSideView(TreeNode *root) {
vector<int> result;
if (root == nullptr)
return result;
queue<bucket> q;
q.push(bucket(root, 0));
int level = 0;
while (!q.empty()) {
auto b = q.front();
q.pop();
auto p = b.node;
if (p->left)
q.push(bucket(p->left, level + 1));
if (p->right)
q.push(bucket(p->right, level + 1));
if (q.empty() || q.front().level != level) {
result.push_back(p->val);
level = q.front().level;
}
}
return result;
}
};