144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    
     2
    /
   3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

M1: recursive

time: O(n), space: O(height)

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

M2: iterative

time: O(n), space: O(n)　　-- depending on the tree structure, could keep up to the entire tree

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