javascript 树结构 - 遍历树结构并显示子项内计数属性的总和

Question

所以我有一个如下所示的树结构，其中叶节点中有一个计数属性。我想总结计数并将计数总和与其父项进行比较。因此parent1和parent2有他们的 child 的总和。然后从那里继续，因此 grantparent 具有 Parent1 和 Parent2 的总和。我也有遍历树的功能。但我无法计数。

有什么想法吗？

javascript

 function transverse(element, result, isSegmentData) {
        if (element instanceof Array){
            element.forEach(function (item) {
                { transverse(item, result, isSegmentData); }
            });
        }
        else if (element instanceof Object) {
            if (element.hasOwnProperty("count")) {

               // sum the count and provide to parent
            }
            if (element.hasOwnProperty("childNodes")) {
                transverse(element.childNodes, result, isSegmentData);
            }
        }
    }

树结构

    [
  {
    "nodeId": 66318,
    "nodeName": "grand parent",
    "childNodes": [
      {
        "nodeId": 66323,
        "nodeName": "parent1",
        "childNodes": [
          {
            "nodeId": 66324,
            "nodeName": "child1",
            "childNodes": [],
            "count": 25
          },
          {
            "nodeId": 66334,
            "nodeName": "child2",
            "childNodes": [],
            "count": 85
          },
          {
            "nodeId": 66439,
            "nodeName": "child3",
            "childNodes": [],
            "count": 65
          },
          {
            "nodeId": 66462,
            "nodeName": "child4",
            "childNodes": [],
            "count": 954
          }
        ]
      },
      {
        "nodeId": 66323,
        "nodeName": "parent2",
        "childNodes": [
          {
            "nodeId": 66324,
            "nodeName": "child1",
            "childNodes": [],
            "count": 225
          },
          {
            "nodeId": 66334,
            "nodeName": "child2",
            "childNodes": [],
            "count": 815
          }
        ]
      }
    ]
  }
]

Answer 1

您可以使用命名函数并再次为数组调用它。然后分配计数，如果没有给出，则从子级中获取计数。

var data = [{ "nodeId": 66318, "nodeName": "grand parent", "childNodes": [{ "nodeId": 66323, "nodeName": "parent1", "childNodes": [{ "nodeId": 66324, "nodeName": "child1", "childNodes": [], "count": 25 }, { "nodeId": 66334, "nodeName": "child2", "childNodes": [], "count": 85 }, { "nodeId": 66439, "nodeName": "child3", "childNodes": [], "count": 65 }, { "nodeId": 66462, "nodeName": "child4", "childNodes": [], "count": 954 }] }, { "nodeId": 66323, "nodeName": "parent2", "childNodes": [{ "nodeId": 66324, "nodeName": "child1", "childNodes": [], "count": 225 }, { "nodeId": 66334, "nodeName": "child2", "childNodes": [], "count": 815 }] }] }];

data.reduce(function x(r, a) {
    a.count = a.count || Array.isArray(a.childNodes) && a.childNodes.reduce(x, 0) || 0;
    return r + a.count;
}, 0);

document.write('<pre>' + JSON.stringify(data, 0, 4) + '</pre>');
console.log(data);

.as-console-wrapper { max-height: 100% !important; top: 0; }