LeetCode(144):Binary Tree Preorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,2,3].

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

非递归法（用栈来还原递归过程）：

public class Solution {
      static  List<Integer> res = new ArrayList<>();
      static  Stack<TreeNode> stack = new Stack<>();
      public static List<Integer> preorderTraversal_inStack(TreeNode root) {
          if(root == null) return new ArrayList<Integer>();
            stack.push(root);
            while(!stack.isEmpty()){
                TreeNode tr = stack.pop();
                res.add(tr.val);
                if(tr.right!=null){
                    stack.push(tr.right);
                }
                if(tr.left!=null){
                    stack.push(tr.left);
                }
            }
            return res; 
        }

递归法：

 public static List<Integer> preorderTraversal(TreeNode root) {
          if(root==null) return new ArrayList<Integer>();
          res.add(root.val);
          while(root.right!=null||root.left!=null){
              preorderTraversal(root.right);
              preorderTraversal(root.left);
          }
          return res;
      }