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?

Recursive：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.ArrayList;
import java.util.List;
public class Solution {
    public void preorder(TreeNode root, List<Integer> list){
	if( root != null ){
			list.add(root.val);
			preorder(root.left,list);
			preorder(root.right,list);
		}
	}
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
		preorder(root, list);
		return list;
    }
}

Iterative：

迭代方法其實就是用一個棧模擬遞歸過程。

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
public class Solution {
	public List<Integer> preorderTraversal(TreeNode root){
		List<Integer> list = new ArrayList<Integer>();
		if( root == null ) 
			return list;
		LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
		while( root != null || !stack.isEmpty() ){
			if( root != null ){
				stack.push(root);
				list.add(root.val);
				root = root.left;
			}else{
				root = stack.pop();
				root = root.right;
			}
		}
		return list;
	}
}