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?

这道题考察用 stack实现iterative前序遍历，代码如下：

/**
 * 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> list = new ArrayList<Integer>();
        Stack<TreeNode> stack =new Stack<TreeNode>();
        if(root == null){
            return list;
        }
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            list.add(node.val);
            if(node.right != null){
                stack.push(node.right);
            }
            if(node.left != null){
                stack.push(node.left);
            }
        }
        return list;
        
    }
}

 Recursive 代码如下：

 public class Solution {     
  
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        if(root == null){
            return list;
        }
　　　　　list.add(root.val);
        list.addAll(postorderTraversal(root.left));
        list.addAll(postorderTraversal(root.right));        
        return list;    
    }

}