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]
]

Solution1：BFS

思路： BFS + 交替每行cur_result 添加到头/尾
Time Complexity: O(N) Space Complexity: O(N)

Solution2：pre-order DFS with level

思路：DFS过程中不同level的结果list 保存到 对应ArrayList<list> result[level]中
Time Complexity: O(N) Space Complexity: O(N) 递归缓存

Solution1 Code：

class Solution1 {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> result = new LinkedList<List<Integer>>();
        
        if(root == null) return result;
        
        boolean order = true;
        queue.offer(root);
        while(!queue.isEmpty()) {
            int num_on_level = queue.size();
            List<Integer> cur_result = new LinkedList<Integer>();
            for(int i = 0; i < num_on_level; i++) {
                TreeNode cur = queue.poll();
                if(cur.left != null) queue.add(cur.left);
                if(cur.right != null) queue.add(cur.right);
                
                // add result to cur_result
                if(order) cur_result.add(cur.val);
                else cur_result.add(0, cur.val);   
            }
            result.add(cur_result);
            order = order ? false : true;
        }
        return result;
    }
}

Solution2 Code：

class Solution2 {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        dfs(result, root, 0);
        return result;
    }
    
    public void dfs(List<List<Integer>> result, TreeNode root, int level) {
        if (root == null) return;
        if (level == result.size()) {
            result.add(new LinkedList<Integer>());
        }
        List<Integer> list = result.get(level);
        if(level % 2 == 0) list.add(root.val);
        else list.add(0, root.val);
           
        dfs(result, root.left, level + 1);
        dfs(result, root.right, level + 1);
    }
}