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,#,#,15,7},

3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

[  [3],  [20,9],  [15,7]]

思路：与二叉树水平序解题思路几乎相同，得到水平序结果之后。再对偶数链表反转就可以。‘

代码例如以下：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    List
     
      
       > list = new ArrayList
       
        
         >();    public List
         
          
           > zigzagLevelOrder(TreeNode root) { dfs(0,root); //每隔1行交换顺序 for(int i = 1; i < list.size(); i = i+2){ List
           
             al = list.get(i); int len = al.size(); //倒序交换 for(int j = 0; j + j < len-1; j++){ int k = al.get(j); al.set(j, al.get(len-1-j)); al.set(len-1-j, k); } } return list; } /** * 中序遍历，依据深度加入list * @param dep 树的深度 * @param root 根节点 */ private void dfs(int dep,TreeNode root){ if(root == null){ return; } List
            
              al;//依据情况得到al值 if(list.size() > dep){ al = list.get(dep); }else{ al = new ArrayList
             
              (); list.add(al); } dfs(dep+1,root.left); al.add(root.val); dfs(dep+1,root.right); }}