Given the root of 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.
Example 1:
Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]
Example 2:
Input: root = [1,null,3]
Output: [1,3]
Example 3:
Input: root = []
Output: []
Constraints:
The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100
类似二叉树的层序遍历,记录每一层中弹出的最后一个数即可
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> res;
queue<TreeNode*> q;
if(root != nullptr) q.push(root);
while(!q.empty()){
int size = q.size();
while(size--){
TreeNode* tmp = q.front();
q.pop();
if(size == 0) res.push_back(tmp->val);
if(tmp->left != nullptr) q.push(tmp->left);
if(tmp->right != nullptr) q.push(tmp->right);
}
}
return res;
}
};