Given a binary tree, return the preorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

分析：也就是求二叉樹的先序遍歷。

思路一：遞迴，DFS求解。比較簡單

 1 class Solution {
 
 2     List<Integer> res = new ArrayList<Integer>();
 3     public List<Integer> preorderTraversal(TreeNode root) {
 4         helper(root);
 5         return res;
 6     }
 7     private void helper(TreeNode root){
 8         if ( root == null ) return ;
 9         res.add(root.val);
 
10         helper(root.left);
11         helper(root.right);
12     }
13 }

思路二：非遞迴方法實現。真正面試如果考到二叉樹的三種遍歷方法，肯定是考非遞迴，也就是棧的方法來實現的。

參考：https://www.cnblogs.com/boris1221/p/9398848.html

 1 class Solution {
 2     public List<Integer> preorderTraversal(TreeNode root) {
 3         List<Integer> list = new
 
 ArrayList<>();
 4         if ( root == null ) return list;
 5         Stack<TreeNode> stack = new Stack<>();
 6         stack.push(root);
 7         while ( !stack.isEmpty() ){
 8             while ( root != null ){
 9                 list.add(root.val);
10                 stack.push(root);
11                 root = root.left;
12             }
13             if ( !stack.isEmpty() ){
14                 root = stack.pop().right;
15             }
16         }
17         return list;
18     }
19 }