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Given a binary tree, return the zigzag level order traversal of its nodes‘ values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ 9 20
/ 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
思路：bfs，偶数行（i%2==1)结果reverse就可以。
/** * 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>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> res= new ArrayList<List<Integer>>(); if(root==null) { return res; } Queue<TreeNode> check=new LinkedList<TreeNode>(); check.offer(root); int level=0; while(!check.isEmpty()) { List<Integer> curl=new ArrayList<Integer>(); for(int i=check.size()-1;i>=0;i--) { TreeNode find=check.poll(); curl.add(find.val); if(find.left!=null) { check.offer(find.left); } if(find.right!=null) { check.offer(find.right); } } if(level%2!=0) { Collections.reverse(curl); } res.add(curl); level++; } return res; } }
103. Binary Tree Zigzag Level Order Traversal
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原文地址：http://www.cnblogs.com/Machelsky/p/5898563.html
