# Binary Tree Level Order Traversal

## Problem

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.

## Code - Java

/**
* 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) {
ArrayList<ArrayList<Integer>> result = new ArrayList<>();
traverse(root, 0, result);
Stack<ArrayList<Integer>> stack = new Stack<ArrayList<Integer>>();
for (ArrayList<Integer> row : result) {
stack.push(row);
}
result = new ArrayList<>();
while (!stack.isEmpty()) {
}
return result;
}

private void traverse(TreeNode node, int level, ArrayList<ArrayList<Integer>> result) {
if (node == null) return;

while (level >= result.size()) {
}