102Binary Tree Level Order Traversal

Posted 2020-06-17

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<List<Integer>> levelOrder(TreeNode root) {
12         List<List<Integer>> returnVal = new ArrayList<List<Integer>>();
13         if (root == null)
14             return returnVal;
15         Queue<TreeNode> nodes = new ArrayDeque<TreeNode>();
16         nodes.offer(root);
17         
18         Queue<TreeNode> nextLevel = new ArrayDeque<TreeNode>();
19         
20         List<Integer> currentList = new ArrayList<Integer>();
21         while(!nodes.isEmpty())
22         {
23             TreeNode next = nodes.poll();
24             currentList.add(next.val);
25             if(next.left != null)
26                 nextLevel.offer(next.left);
27             if(next.right != null)
28                 nextLevel.offer(next.right);
29             
30             if(nodes.isEmpty())
31             {
32                 nodes = nextLevel;
33                 nextLevel = new ArrayDeque<TreeNode>();
34                 returnVal.add(currentList);
35                 currentList = new ArrayList<Integer>();
36             }
37         }
38         
39         
40         return returnVal;
41     }
42 }