**Problem:**

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:**

BFS

/** * 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>> result = new ArrayList<List<Integer>>(); if (root == null) return result; Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); while (!q.isEmpty()) { int size = q.size(); List<Integer> row = new ArrayList<Integer>(); while (size > 0) { TreeNode current = q.poll(); row.add(current.val); if (current.left != null) q.offer(current.left); if (current.right != null) q.offer(current.right); size--; } result.add(row); } return result; } }