Binary Tree Level Order Traversal

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],

  1.     3
  2.    / \
  3.   9  20
  4.     /  \
  5.    15   7

return its level order traversal as:

  1. [
  2.   [3],
  3.   [9,20],
  4.   [15,7]
  5. ]

Solution:

  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>> res = new ArrayList<List<Integer>>();
  14.     if (root == null) {
  15.       return res;
  16.     }
  18.     // Use a queue to help level order traversal
  19.     Queue<TreeNode> queue = new LinkedList<TreeNode>();
  21.     queue.offer(root);
  22.     queue.offer(null);
  24.     List<Integer> list = new ArrayList<Integer>();
  26.     while (!queue.isEmpty()) {
  27.       TreeNode node = queue.poll();
  29.       if (node != null) {
  30.         list.add(node.val);
  32.         if (node.left != null) {
  33.           queue.offer(node.left);
  34.         }
  36.         if (node.right != null) {
  37.           queue.offer(node.right);
  38.         }
  39.       } else {
  40.         res.add(new ArrayList<Integer>(list));
  41.         list = new ArrayList<Integer>();
  43.         if (!queue.isEmpty()) {
  44.           queue.offer(null);
  45.         }
  46.       }
  47.     }
  49.     return res;
  50.   }
  51. }