[LeetCode] 102. Binary Tree Level Order Traversal Java
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题目:
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] ]
题意及分析:宽度遍历一棵树。用队列保存中间值即可。
代码:
/** * 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>> res = new ArrayList<>(); if(root==null) return res; List<Integer> list = new ArrayList<>(); Queue<TreeNode> queue=new LinkedList<>(); queue.offer(root); Queue<TreeNode> queue1=new LinkedList<>(); while(!queue.isEmpty()||!queue1.isEmpty()){ queue1.clear(); while(!queue.isEmpty()){ //遍历一层 TreeNode now = queue.poll(); list.add(now.val); if(now.left!=null) queue1.offer(now.left); if(now.right!=null) queue1.offer(now.right); } queue=new LinkedList<>(queue1); //进行下一层的遍历 res.add(new ArrayList<>(list)); list.clear(); //清空 } return res; } }
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