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.
Solution:
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 List<List<Integer>> levelOrder(TreeNode root) { 12 List<List<Integer>> res = new ArrayList<List<Integer>>(); 13 List<Integer> oneRes = new ArrayList<Integer>(); 14 if (root==null){ 15 return res; 16 } 17 18 List<TreeNode> lastLevel = new ArrayList<TreeNode>(); 19 lastLevel.add(root); 20 oneRes.add(root.val); 21 res.add(oneRes); 22 List<TreeNode> curLevel; 23 while (lastLevel.size()!=0){ 24 oneRes = new ArrayList<Integer>(); 25 curLevel = new ArrayList<TreeNode>(); 26 for (int i=0;i<lastLevel.size();i++){ 27 TreeNode curNode = lastLevel.get(i); 28 if (curNode.left!=null){ 29 curLevel.add(curNode.left); 30 oneRes.add(curNode.left.val); 31 } 32 if (curNode.right!=null){ 33 curLevel.add(curNode.right); 34 oneRes.add(curNode.right.val); 35 } 36 } 37 if (curLevel.size()==0){ 38 break; 39 } else { 40 res.add(oneRes); 41 lastLevel = curLevel; 42 } 43 } 44 45 return res; 46 } 47 }