[Leetcode] 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?

Solution 1: 非递归

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result=new ArrayList<Integer>();   
        if(root==null){
            return result;
        }
        Stack<TreeNode> s=new Stack<TreeNode>();
        result.add(root.val);
        s.add(root);
        TreeNode p=root.left;
        
        while(!s.isEmpty()){
            while(p!=null){
                s.add(p);
                result.add(p.val);
                p=p.left;
            }
            TreeNode n=s.pop();
            p=n.right;
            if(p!=null){
                s.add(p);
                result.add(p.val);
                p=p.left;
            }
        }
        return result;
    }
}

Solution 2: 递归

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result=new ArrayList<Integer>();
        myPreorderTraversal(root,result);
        return result;
    }

    private void myPreorderTraversal(TreeNode root, List<Integer> result) {
        // TODO Auto-generated method stub
        if(root!=null){
            result.add(root.val);
            myPreorderTraversal(root.left, result);
            myPreorderTraversal(root.right, result);
        }
    }
}