Binary Tree Level Order Traversal.java

/*
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,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.


OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:
   1
  / \
 2   3
    /
   4
    \
     5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

*/

public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if (root == null) {
            return result;
        }
        
        ArrayList<TreeNode> currentList = new ArrayList<TreeNode>();
        currentList.add(root);
       
        
        while (currentList.size() != 0) {
            ArrayList<TreeNode> nextLevelList = new ArrayList<TreeNode>();
            ArrayList<Integer> parentLevelData = new ArrayList<Integer>();
            
            for (TreeNode s : currentList) {
                parentLevelData.add(s.val);
                if (s.left != null) {
                    nextLevelList.add(s.left);
                }
                if (s.right != null) {
                    nextLevelList.add(s.right);
                }
            }
            result.add(parentLevelData);
            //currentList = new ArrayList<TreeNode>(nextLevelList);
            currentList = nextLevelList;
        }
        
        return result;
    }
}