199. Binary Tree Right Side View

# Medium

Too easy. BFS + root-right-left traversal. On each layer, first element in queue is the right side view.

C++

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        // edge case
        vector<int> result;
        if(root == NULL) return result;
        if(root->left == NULL and root->right == NULL) {
            result.push_back(root->val);
            return result;
        }
        
        // initialize
        
        vector<TreeNode*> queue;
        queue.push_back(root);
        
        // regular case, BFS
        while(queue.size() != 0) {
            int l = queue.size();
            for(int i = 0; i < l; i ++) {
                root = queue.at(0);
                queue.erase(queue.begin());
                if(i == 0) result.push_back(root->val);                
                if(root->right != NULL) queue.push_back(root->right);
                if(root->left != NULL) queue.push_back(root->left);
            }
        }
        
        return result;
    }
};

Time complexity = $O(n)$ , space complexity = $O(n)$ , 这里的空间复杂度是额外定义的队的大小，最大为 $n$ . $n$是总共的节点数。

Second Tab