如何删除/修剪节点中不存在于单独数组中的所有树节点
我有以下 JSON 树表示形式：
let tree = [
{
"label": "Org01",
"children": [
{
"label": "Dist01",
"children": [
{
"label": "School1",
"children": [
{
"label": "class1",
"children": []
},
{
"label": "class2",
"children": []
}
]
},
{
"label": "School1tst",
"children": []
}
]
},
{
"label": "Dist02",
"children": []
}
]
},
{
"label": "contoso01",
"children": [
{
"label": "Dist A",
"children": [
{
"label": "School A",
"children": [
{
"label": "classA",
"children": []
}
]
},
{
"label": "School B",
"children": [
{
"label": "classB",
"children": []
}
]
}
]
}
我有一个数组中的节点列表，如下所示：
let whitelist = ['class1', 'School1', 'Dist01'];
如何从树中删除上述数组中不存在的所有节点。但是，如果父节点的子节点在白名单中，则需要在树上显示父节点。
从树中删除特定节点对我来说是可能的，但是除了数组中的少数节点之外，我无法找到从树中删除所有节点的方法。
谢谢，我感谢任何帮助。
慕标58322723回答
-
陪伴而非守候
这应该可以完成工作：function filter(tree, list){ let output = []; for(i in tree){ if(list.indexOf(tree[i].label) >= 0){ tree[i].children = filter(tree[i].children, list); output.push(tree[i]); }else{ output = output.concat(filter(tree[i].children, list)); } } return output;} -
小唯快跑啊
我这样解决了这个问题：function deleteNodes(tree, list) { if (tree.length > 0) { tree.forEach((node, i) => { this.deleteNodes(node.subItems, list); if (node.subItems) { if (node.subItems.length === 0 && !list.includes(node.text)) { tree.splice(i, 1); } } }); }} -
MM们
const prunedNode = node => { const pruned = whitelist.includes(node.label) ? node : null; if (pruned) { node.children = node.children.reduce((prunedChildren, child) => { const prunedChildNode = prunedNode(child); if (prunedChildNode) { prunedChildren.push(prunedChildNode); } return prunedChildren; }, []); } return pruned;};console.log(prunedNode(tree));
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相关分类
JavaScript