首页 > 代码库 > 102. Binary Tree Level Order Traversal
102. 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]
,
3 / 9 20 / 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
层序遍历
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> re = new LinkedList<List<Integer>>(); if (root == null) { return re; } ArrayDeque<TreeNode> queue = new ArrayDeque<TreeNode>(); TreeNode tmp; int size = 0; queue.offer(root); while (!queue.isEmpty()) { List<Integer> l = new LinkedList<Integer>(); size = queue.size(); for (int i = 0; i < size; i++) { tmp = queue.poll(); if (tmp.left != null) { queue.offer(tmp.left); } if (tmp.right != null) { queue.offer(tmp.right); } l.add(tmp.val); } re.add(l); } return re; } }
102. Binary Tree Level Order Traversal
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。