144. Binary Tree Preorder Traversal (Medium)

Given a binary tree, return thepreordertraversal of its nodes' values.

For example:
Given binary tree[1,null,2,3],

   1
    \
     2
    /
   3

return[1,2,3].

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

Solution 1: Recursive, DFS O(n); O(h)

    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        dfs(res, root);
        return res;
    }

    private void dfs(List<Integer> res, TreeNode root) {
        if (root == null) {
            return;
        }

        res.add(root.val);
        dfs(res, root.left);
        dfs(res, root.right);
    }

Solution 2: Iterative + Stack O(n); O(n)不太准确

    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }

        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.empty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) {
                stack.push(cur.right); //FILO, first push right child into stack so that right child can last pop
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }
        }
        return res;
    }