MongoDB:如何通过条件递归获取所有子节点的计数？

// 创建节点集合
db.nodes.insertMany([
  { _id: 1, name: "Node 1", parent: null },
  { _id: 2, name: "Node 2", parent: 1 },
  { _id: 3, name: "Node 3", parent: 1 },
  { _id: 4, name: "Node 4", parent: 2 },
  { _id: 5, name: "Node 5", parent: 2 },
  { _id: 6, name: "Node 6", parent: 3 },
]);

// 递归查询函数
function countChildNodes(nodeId) {
  var count = 0;
  
  // 查询当前节点的子节点
  var children = db.nodes.find({ parent: nodeId });
  
  // 遍历子节点
  children.forEach(function(child) {
    count++; // 增加子节点计数
    
    // 递归查询子节点的子节点
    count += countChildNodes(child._id);
  });
  
  return count;
}

// 获取根节点的子节点计数
var rootNodeId = 1;
var childNodeCount = countChildNodes(rootNodeId);
print("子节点计数: " + childNodeCount);