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,null,null,15,7]`

,

3 / \ 9 20 / \ 15 7

return its level order traversal as:

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

**Solution:**

**Approach 1: Using level order traversal**

public class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> result = new ArrayList<>(); Queue<TreeNode> q = new LinkedList<> (); if(root==null) return result; q.add(root); while(!q.isEmpty()){ List<Integer> subresult = new ArrayList<>(); int n = q.size(); for(int i=0; i<n;i++){ TreeNode node = q.poll(); subresult.add(node.val); if(node.left!=null)q.offer(node.left); if(node.right!=null)q.offer(node.right); } result.add(subresult); } return result; } }

**Approach 2:**

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { List<List<Integer>> ans; public List<List<Integer>> levelOrder(TreeNode root) { ans = new ArrayList<List<Integer>>(); travel(0,root); return ans; } private void travel(int level,TreeNode cur){ if(cur == null) return; else{ if(level >= ans.size()){ List<Integer> list = new ArrayList<Integer>(); ans.add(list); } List<Integer> valList = ans.get(level); valList.add(cur.val); travel(level+1,cur.left); travel(level+1,cur.right); } } }