/*
        if (root == null) return ;
1. 看下递归的写法，每次把左子树节点压入栈中
        recursive(root.left, list);
2. 左子节点全入栈后输出栈顶节点
        list.add(root.val);
3. 把当前节点的右子节点压入栈中
        recursive(root.right, list);
4. 当前节点弹出栈

*/

/**
 * 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> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        iteratively(root, list);
        return list;
    }
    public void iteratively(TreeNode root, List<Integer> list){

        Stack<TreeNode> stack = new Stack<>();
        var p = root;

        // while(p != null && !stack.empty()){
        while(p != null || !stack.empty()){
            while (p != null){
                stack.add(p);
                p = p.left;
            }

            p = stack.pop();
            list.add(p.val);
            p = p.right;
        }
        return ;
    }
    public void iterativelyError(TreeNode root, List<Integer> list){

        Stack<TreeNode> stack = new Stack<>();
        stack.add(root);

        // 这样不ok，因为java只能操作指针的值，不能把对应地址置为null；
        // 即使可以也破坏了原树结构
        while(!stack.empty()){
            var node = stack.peek();

            if (node.left != null) stack.add(node.left);
            else {

                list.add(node.val);
                stack.pop();
                if (node.right != null) stack.add(node.right);
                node = null;
            }
        }
        return ;
    }

    public void recursive(TreeNode root, List<Integer> list){
        if (root == null) return ;
        recursive(root.left, list);
        list.add(root.val);
        recursive(root.right, list);
    }
}