原題傳送門

1. 題目描述

2. Solution 1

1、思路分析
使用BFS+遞迴，逐層遍歷樹，若當前層level為奇數，頭插結點；若當前level為偶數，尾插結點。

2、程式碼實現

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為偶數，尾插
        else curLevel.add(0, cur.val);      // level為奇數，頭插
        travel(cur.left, result, level + 1);
        travel(cur.right, result, level + 1);
    }
}

/*
   中文社群，另解: 1. 正常層次遍歷；2. 遍歷1. 結果, level為奇數，reverse.
 */
 

3、複雜度分析
時間複雜度: O(n)
空間複雜度: O(n)