原题传送门

1. 题目描述

2. Solution 1

1、思路分析
使用BFS+递归，逐层遍历树，若当前层level为奇数，头插结点；若当前level为偶数，尾插结点。

2、代码实现

package Q0199.Q0103BinaryTreeZigzagLevelOrderTraversal;

import DataStructure.TreeNode;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;

public class Solution {
    /*
       1. O(n) solution by using LinkedList along with ArrayList.
        So insertion in the inner list and outer list are both O(1),
       2. Using BFS and creating new lists when needed.
     */
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        travel(root, result, 0);
        return result;
    }

    private void travel(TreeNode cur, List<List<Integer>> result, int level) {
        if (cur == null) return;
        if (result.size() <= level) result.add(new LinkedList<>());

        List<Integer> curLevel = result.get(level);
        if (level % 2 == 0) curLevel.add(cur.val);      // level为偶数，尾插
        else curLevel.add(0, cur.val);      // level为奇数，头插
        travel(cur.left, result, level + 1);
        travel(cur.right, result, level + 1);
    }
}

/*
   中文社区，另解: 1. 正常层次遍历；2. 遍历1. 结果, level为奇数，reverse.
 */

3、复杂度分析
时间复杂度: O(n)
空间复杂度: O(n)

标签：Binary,cur,level,List,travel,Level,Tree,result,import
来源： https://www.cnblogs.com/junstat/p/16252055.html