• Time:O(n)
• Space:O(n)

## C++

``````class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
if (!root)
return {};

vector<vector<int>> ans;
queue<TreeNode*> q{{root}};

while (!q.empty()) {
vector<int> currLevel;
for (int sz = q.size(); sz > 0; --sz) {
TreeNode* node = q.front();
q.pop();
currLevel.push_back(node->val);
if (node->left)
q.push(node->left);
if (node->right)
q.push(node->right);
}
ans.push_back(currLevel);
}

reverse(begin(ans), end(ans));
return ans;
}
};
``````

## JAVA

``````class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
if (root == null)
return new ArrayList<>();

List<List<Integer>> ans = new ArrayList<>();
Queue<TreeNode> q = new ArrayDeque<>(Arrays.asList(root));

while (!q.isEmpty()) {
List<Integer> currLevel = new ArrayList<>();
for (int sz = q.size(); sz > 0; --sz) {
TreeNode node = q.poll();
if (node.left != null)
q.offer(node.left);
if (node.right != null)
q.offer(node.right);
}
}

Collections.reverse(ans);
return ans;
}
}
``````

## Python

``````class Solution:
def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
if not root:
return []

ans = []
q = deque([root])

while q:
currLevel = []
for _ in range(len(q)):
node = q.popleft()
currLevel.append(node.val)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
ans.append(currLevel)

return ans[::-1]
``````