Binary Tree Level Order Traversal Given the root
of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).
Example 1:


Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]]
Example 2:
Input: root = [1] Output: [[1]]
Example 3:
Input: root = [] Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]
. -1000 <= Node.val <= 1000
Binary Tree Level Order Traversal Solutions
✅Time: O(n)
✅Space: O(n)
C++
class Solution {
public:
vector<vector<int>> levelOrder(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);
}
return ans;
}
};
Java
class Solution {
public List<List<Integer>> levelOrder(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();
currLevel.add(node.val);
if (node.left != null)
q.offer(node.left);
if (node.right != null)
q.offer(node.right);
}
ans.add(currLevel);
}
return ans;
}
}
Python
class Solution:
def levelOrder(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
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