Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

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  3
 / \
9  20
  /  \
 15   7

return its level order traversal as:

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5
[
  [3],
  [9,20],
  [15,7]
]

BFS

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>(); // result list
        List<Integer> sub; // sub list for each level
        if (root == null){return res;}
        
        Queue<TreeNode> level = new LinkedList<>();
        level.add(root);
        
        while (level.size() > 0){
            sub = new ArrayList<>();
            int levelNum = level.size();
            
            for (int i = 0; i < levelNum; i++){
                TreeNode cur = level.poll();
                if (cur.left != null) level.add(cur.left);
                if (cur.right != null) level.add(cur.right);
                sub.add(cur.val);
            }
            res.add(sub);
        }
        return res;
    }
}

DFS

```java
public class Solution {
public List> levelOrder(TreeNode root) {
List> res = new ArrayList>();
levelHelper(res, root, 0);
return res;
}

public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
    if (root == null) return;
    if (height >= res.size()) {
        res.add(new LinkedList<Integer>());
    }
    res.get(height).add(root.val);
    levelHelper(res, root.left, height+1);
    levelHelper(res, root.right, height+1);
}

}
```A