Java [Leetcode 102]Binary Tree Level Order Traversal
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题目描述:
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] ]
解题思路:
运用广度优先搜索方法,同http://www.cnblogs.com/zihaowang/p/5149745.html类似。
代码如下:
/** * 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>> result = new LinkedList<List<Integer>>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); if(root == null) return result; queue.offer(root); while(!queue.isEmpty()){ int m = queue.size(); List<Integer> list = new LinkedList<Integer>(); for(int i = 0; i < m; i++){ if(queue.peek().left != null) queue.offer(queue.peek().left); if(queue.peek().right != null) queue.offer(queue.peek().right); list.add(queue.poll().val); } result.add(new LinkedList<Integer>(list)); } return result; } }
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