标签：init order java 广度 mil eve amp family roo
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] ]
广度优先搜索，一般做法都是采取一个队列来做。
不过这题有个比较不一样的是他返回的格式是List<List>>的格式，所以需要做一些处理。
直接上代码了，很简单的题，
JAVA CODE:
/** * 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>> returnList = new ArrayList<List<Integer>>(); if(root == null){ return returnList; } List<Integer> temp = null; Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.add( root ); while( !queue.isEmpty() ) { Queue<TreeNode> tempQ = new LinkedList<TreeNode>(); temp = new ArrayList<Integer>(); while( !queue.isEmpty() ) { TreeNode tn = queue.poll(); if( tn.left != null ) { tempQ.add( tn.left ); } if( tn.right != null ) { tempQ.add( tn.right ); } temp.add( tn.val ); } queue = tempQ; returnList.add( temp ); } return returnList; } }
【LeetCode】Binary Tree Level Order Traversal 【BFS】
标签：init order java 广度 mil eve amp family roo
原文地址：http://www.cnblogs.com/dick159/p/6508855.html
