# ac solution code

• ``````/*
Solution1. time = O(n); space = O(n)
The basic idea is using BFS to traverse the tree level by level. The little trick is to get the current count of the queue first, then loop though the current level up to count, so that we can reuse the queue.
*/
func levelOrder(_ root: TreeNode?) -> [[Int]] {
var res = [[Int]]()
guard let root = root else {return res}
var queue = Queue<TreeNode>()
queue.offer(root)

while !queue.isEmpty() {            // BFS loop
let count = queue.size
var oneLevel = [Int]()
for _ in 0 ..< count {          // Loop though the current level up to count to reuse queue
let node = queue.poll()!    // Dequeue the first node
oneLevel.append(node.val)   // Append node's val to oneLevel
if let left = node.left {   // Enqueue node's left
queue.offer(left)
}
if let right = node.right { // Enqueue node's right
queue.offer(right)
}
}
res.append([Int](oneLevel))     // Append oneLevel to res
}
return res
}
``````

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