递归中的变量被重置。有人能解释一下吗？

Question

我试图从Leetcode解决这个问题https://leetcode.com/problems/count-good-nodes-in-binary-tree/

这是我的解决方案：我无法理解为什么这个递归是来自根的计数值。左节点在遍历根时不起作用。正当据我所知，我是

1. 检查当前节点是否良好并更新计数和列表
2. 遍历左节点更新计数
3. 上面的count应该在遍历右节点时进入右节点并更新计数值，但没有发生

为什么这不起作用。我知道正确的解决方案，只是不能真正理解递归如何重置我的计数变量

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode() {}
     *     TreeNode(int val) { this.val = val; }
     *     TreeNode(int val, TreeNode left, TreeNode right) {
     *         this.val = val;
     *         this.left = left;
     *         this.right = right;
     *     }
     * }
     */
    class Solution {
        public int goodNodes(TreeNode root) {
            int count = 0;
            List<Integer> list = new ArrayList<>();
            TreeNode treeroot = root;
            preorderGoodNodes(root,list,count,treeroot);
            return count; 
        }
        
       public void  preorderGoodNodes(TreeNode root,List<Integer> list,int count,TreeNode treeroot)
       {
        
           if(root==null) // check if root is null
               return;
           
           // if current node is actual root of the tree then count ++ since root is always good
           //also add the root to the list
           if(treeroot ==root) 
           {
               count++;
               list.add(root.val);
           }
           else
               // if node is not the root then check if it is good or not by looking into the list
           {  
                   int flag = 0;
                   for(int x : list) //looking into the list
                   {
                       if(x>root.val)
                       {
                           flag = 1;
                           break;
                       }
    
                   }
    
                   if(flag==0) // if it is good count++
                           count++;
                       
               list.add(root.val); // weather good or not add to the list
    
           }
           
           List<Integer> rightlist = new ArrayList<>(list); 
           // make a copy of the list to send to right node 
           //because count and list from left tree should not effect right tree
           
           preorderGoodNodes(root.left,list,count,treeroot);
**// why does count reset after this line ??**
           preorderGoodNodes(root.right,rightlist,count,treeroot);
           
       }
    }

Answer 1

如果需要“通过引用传递”语义，可以使用大小为1的AtomicInteger或int[]而不是int。