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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.
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 5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}"
.思路:利用队列进行处理就行了,每次记录入队的有几个,就是下一层要访问的了
/** * Definition for binary tree * 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>> result = new ArrayList<List<Integer>>(); result.clear(); if (root == null) return result; LinkedList<TreeNode> q = new LinkedList<TreeNode>(); q.add(root); int level = 0, count = 1; while (!q.isEmpty()) { List<Integer> tmp = new ArrayList<Integer>(); tmp.clear(); level = 0; for (int i = 0; i < count; i++) { root = q.pollFirst(); tmp.add(root.val); if (root.left != null) { q.add(root.left); ++level; } if (root.right != null) { q.add(root.right); ++level; } } count = level; result.add(tmp); } return result; } }
LeetCode Binary Tree Level Order Traversal
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原文地址:http://blog.csdn.net/u011345136/article/details/45347741