103. Binary Tree Zigzag Level Order Traversal

Medium
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]
]

这道题据说是面试常考题，这一类基础的高频题要务必掌握。我用的方法比较简单，就用BFS按照Level Order Traversal来遍历。然后选择奇数层，翻转一下对应的ArrayList<Integer> 就可以了。翻转的写法我是搜到的，list.add(i,e)这个写法是把Element e insert到index = i的位置，很少用但很方便的一个method.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null){
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()){
            int size = queue.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i++){
                TreeNode curt = queue.poll();
                level.add(curt.val);
                if (curt.left != null){
                    queue.offer(curt.left);
                }
                if (curt.right != null){
                    queue.offer(curt.right);
                }
            }
            res.add(level);
        }
        
        for (int i = 0; i < res.size(); i++){
            if (i % 2 != 0){
                List<Integer> list = res.get(i);
                for (int j = 0, k = list.size() - 1; j < k; j++){
                    list.add(j, list.remove(k));    
                }  
            }
        }
        return res;
    }
}