Binary Tree Preorder Traversal

原题链接
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?

思路解析

• 非递归遍历通常都是使用栈来完成，（递归的本质也是栈）前序遍历二叉树，先访问节点，然后把右子树入栈，再把左子树入栈， 如此循环即可

 public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if (root == null) {
            return list;
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()) {
            root = stack.pop();
            list.add(root.val);
            if (root.right != null) {
                stack.push(root.right);
            }
            if (root.left != null) {
                stack.push(root.left);
            }
        }
        return list;
    }