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?

如果不能使用recursive 那么就只能使用stack来保存状态。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        Stack<TreeNode> st = new Stack<TreeNode>();
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root == null) return result;
        st.push(root);
        while(!st.isEmpty()){
            TreeNode tmp = st.pop();
            result.add(tmp.val);
            if(tmp.right != null) st.push(tmp.right);
            if(tmp.left != null) st.push(tmp.left);
        }
        return result;
    }
}

注意要现存right，再存left，这样才能保证先expand的是left。