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Leetcode-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,#,#,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 }
Leetcode-Binary Tree Level Order Traversal
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