LeetCode Solutions
199. Binary Tree Right Side View
Time: $O(n)$ Space: $O(n)$
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
if (root == nullptr)
return {};
vector<int> ans;
queue<TreeNode*> q{{root}};
while (!q.empty()) {
const int size = q.size();
for (int i = 0; i < size; ++i) {
TreeNode* node = q.front();
q.pop();
if (i == size - 1)
ans.push_back(node->val);
if (node->left)
q.push(node->left);
if (node->right)
q.push(node->right);
}
}
return ans;
}
};
class Solution {
public List<Integer> rightSideView(TreeNode root) {
if (root == null)
return new ArrayList<>();
List<Integer> ans = new ArrayList<>();
Queue<TreeNode> q = new ArrayDeque<>(Arrays.asList(root));
while (!q.isEmpty()) {
final int size = q.size();
for (int i = 0; i < size; ++i) {
TreeNode node = q.poll();
if (i == size - 1)
ans.add(node.val);
if (node.left != null)
q.offer(node.left);
if (node.right != null)
q.offer(node.right);
}
}
return ans;
}
}
class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
if not root:
return []
ans = []
q = deque([root])
while q:
size = len(q)
for i in range(size):
root = q.popleft()
if i == size - 1:
ans.append(root.val)
if root.left:
q.append(root.left)
if root.right:
q.append(root.right)
return ans