leetCode 103.Binary Tree Zigzag Level Order Traversal （二叉树Z字形水平序） 解题思路和方法

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<Integer>> list = new ArrayList<List<Integer>>();
    public List<List<Integer>> 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<Integer> 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<Integer> al;//依据情况得到al值
        if(list.size() > dep){
            al = list.get(dep);
        }else{
            al = new ArrayList<Integer>();
            list.add(al);
        }
        dfs(dep+1,root.left);
        al.add(root.val);
        dfs(dep+1,root.right);
    }
}