Level Traversal

//BFS need queue
public class Solution {
    /**
     * @param root: A Tree
     * @return: Level order a list of lists of integer
     */
    public List<List<Integer>> levelOrder(TreeNode root) {
        // write your code here
        
        Queue<TreeNode> queue = new LinkedList<>();
        
        List<List<Integer>> res = new ArrayList<>();
        
        if(root == null){
            return res;
        }
        
        queue.offer(root);
        
        while(!queue.isEmpty()){
            int size = queue.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < size ;++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);
                }
            } 
            
            res.add(level);
        }
        
        return res;
    }
}

//dfs O(n)
public class Solution {
    /**
     * @param root: A Tree
     * @return: Level order a list of lists of integer
     */
    public List<List<Integer>> levelOrder(TreeNode root) {
        // write your code here
        List<List<Integer>> result = new ArrayList<>();
        
        if (root == null) {
            return result;
        }
        
        dfs(root, result, 1);
        
        return result;
    }
    
    private void dfs(TreeNode node, List<List<Integer>> result, int level) {
        if (node == null) {
            return;
        }
        
        if (result.size() < level) {
            List<Integer> list = new ArrayList<Integer>();
            list.add(node.val);
            result.add(list);
        } else {
            result.get(level - 1).add(node.val);
        }
        
        dfs(node.left, result, level + 1);
        dfs(node.right, result, level + 1);
    }
}