# Binary Tree Level Order Traversal

Easy

## Question

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

Example
Given binary tree {3,9,20,#,#,15,7} ,

``````    3
/ \
9  20
/  \
15   7
``````

return its level order traversal as:

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

Challenge
Challenge 1: Using only 1 queue to implement it.
Challenge 2: Use DFS algorithm to do it.

## Solution

#### Java (Queue)

``````/**
* Definition of TreeNode:
* public class TreeNode {
*     public int val;
*     public TreeNode left, right;
*     public TreeNode(int val) {
*         this.val = val;
*         this.left = this.right = null;
*     }
* }
*/

public class Solution {
/**
* @param root: The root of binary tree.
* @return: Level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
// write your code here
Queue<TreeNode> queue = new LinkedList<TreeNode>();
ArrayList<ArrayList<Integer>> totalList = new ArrayList<ArrayList<Integer>>();
if (root == null) {
return totalList;
}
queue.offer(root);
while (!queue.isEmpty()) {
int levelsize = queue.size();
ArrayList<Integer> list = new ArrayList<Integer>();
for (int i = 0; i < levelsize; i++) {
TreeNode node = queue.poll();
list.add(node.val);
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
totalList.add(list);
}
return totalList;
}
}
``````