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]]
The number of nodes in the tree is in the range [0, 2000].BFS with a queue. At each level, record queue size to know when the level ends.
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) return result;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int levelSize = queue.size(); // snapshot size before processing
List<Integer> level = new ArrayList<>();
for (int i = 0; i < levelSize; i++) {
TreeNode node = queue.poll();
level.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
result.add(level);
}
return result;
}Time: O(n) · Space: O(n)