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 – Easy to understand

/** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { public IList<IList<int>> LevelOrder(TreeNode root) { var result = new List<IList<int>>(); if(root == null) return result; Queue<TreeNode> q = new Queue<TreeNode>(); q.Enqueue(root); while(q.Count > 0) { List<int> list = new List<int>(); int size = q.Count; for(int i=0;i<size;i++) { TreeNode node = q.Dequeue(); list.Add(node.val); if(node.left != null) q.Enqueue(node.left); if(node.right != null) q.Enqueue(node.right); } result.Add(list); } return result; } }

Time Complexity: O(n)

Space Complexity: O(n)