{
platformId: 3,
areaId: "<ROOT>",
areaName: "ROOT",
parentAreaId: "",
online: 0,
offline: 0,
total: 0,
sum: 0,
sumOnline: 0,
sumOffline: 0,
children: [{
platformId: 3,
areaId: "000",
areaName: "测试",
parentAreaId: "<ROOT>",
online: 2,
offline: 1,
total: 3,
sum: 0,
sumOnline: 0,
sumOffline: 0,
children: [{
platformId: 3,
areaId: "000002",
areaName: "测试2",
parentAreaId: "000",
online: 0,
offline: 0,
total: 0,
sum: 0,
sumOnline: 0,
sumOffline: 0,
children: []
},
{
platformId: 3,
areaId: "000001",
areaName: "测试1",
parentAreaId: "000",
online: 30,
offline: 0,
total: 30,
sum: 0,
sumOnline: 0,
sumOffline: 0,
children: []
},
{
platformId: 3,
areaId: "000000",
areaName: "测试0",
parentAreaId: "000",
online: 274,
offline: 10,
total: 284,
sum: 0,
sumOnline: 0,
sumOffline: 0,
children: []
}
]
}]
}这种树结构的数据怎么遍历统计，每个结点的sum要是下级节点total的总和及本节点total的总和，每个节点的sumoffline要是下级节点及本节点offline的总和
private AreaCameraStatusSum count(AreaCameraStatusSum root){
int total = root.getTotal();
int offline = root.getOffline();
if(root.getChildren().size() < 1){
root.setSum(total);
root.setSumOffline(offline);
return root;
}
int sum = total;
int sumOffline = offline;
for(AreaCameraStatusSum child : root.getChildren()){
AreaCameraStatusSum tmp = count(child);
sum += tmp.getSum();
sumOffline += tmp.getSumOffline();
}
root.setSum(sum);
root.setSumOffline(sumOffline);
return root;
}上面用递归的方法可以求出每个节点的和，不知道有没有更好的方法
