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?

前序遍历用递归很容易写，用迭代需要用到堆栈。代码如下：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> preorderTraversal(TreeNode root) {        Stack<TreeNode> stack = new Stack<TreeNode>();        List<Integer> res = new ArrayList<Integer>();        while (root != null) {            res.add(root.val);            if (root.right != null) {                stack.push(root.right);            }            root = root.left;            if (root == null && !stack.isEmpty()) {                root = stack.pop();            }        }        return res;    }}