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?

Analysis: 第一反应肯定是recursion(递归), 非常直接，但是题目要求不能用递归。如果要使用迭代的方法来写preorder traversal，最大的问题是要如何确定遍历节点的顺序，因为树的pre-order traversal其实很类似图的DFS，DFS可以用Stack来写，所以这里写pre-order traversal也可以用stack来实现迭代的写法。

Iterative: 其实就是用一个栈来模拟递归的过程。所以算法时间复杂度也是O(n)，空间复杂度是栈的大小O(logn)。

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> preorderTraversal(TreeNode root) {
12         List<Integer> res = new ArrayList<>();
13         TreeNode p = root;
14         Stack<TreeNode> stack = new Stack<>();
15         while (p!=null || !stack.isEmpty()) {
16             if (p != null) {
17                 stack.push(p);
18                 res.add(p.val);
19                 p = p.left;
20             }
21             else {
22                 TreeNode node = stack.pop();
23                 p = node.right;
24             }
25         }
26         return res;
27     }
28 }

recursion:

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public ArrayList<Integer> preorderTraversal(TreeNode root) {
12         ArrayList<Integer> result = new ArrayList<Integer>();
13         preorder(root, result);
14         return result;
15     }
16     
17     public void preorder(TreeNode root, ArrayList<Integer> result) {
18         if (root == null) return;
19         result.add(root.val);
20         preorder(root.left, result);
21         preorder(root.right, result);
22     }
23 }