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}".1 /** 2 * Definition for binary tree 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 ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) { 12 // IMPORTANT: Please reset any member data you declared, as 13 // the same Solution instance will be reused for each test case. 14 ArrayList<TreeNode> queue = new ArrayList<TreeNode>(); 15 ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); 16 if(root == null) return result; 17 queue.add(root); 18 TreeNode nl = new TreeNode(Integer.MIN_VALUE); 19 queue.add(nl); 20 ArrayList<Integer> cur = new ArrayList<Integer>(); 21 while(queue.size() != 1){ 22 TreeNode rn = queue.remove(0); 23 if(rn.val == Integer.MIN_VALUE){ 24 queue.add(nl); 25 result.add(cur); 26 cur = new ArrayList<Integer>(); 27 } 28 else{ 29 if(rn.left != null) queue.add(rn.left); 30 if(rn.right != null) queue.add(rn.right); 31 cur.add(rn.val); 32 } 33 } 34 result.add(cur); 35 return result; 36 } 37 }