引言
最近接手一个项目中很多地方需要用到树形结构表格等,因此自己封了个VUE的树和表格组件,需要经常对两种形式的数据进行相互转换,这里记录下转换的方法,组件等有时间再完善下也发上来。
扁平数组转换为树形结构
这个是最常用的,当我们从后台获取一个扁平数组的时候,通常比如用id
、pid
来标识父子关系,如:
var arr = [{id: 1, pid: '-1'},{id: 11, pid: '1'},{id: 12, pid: '1'}]
用map
记录的方法是最常用效果也最好的复杂度是O(nlgn)
,支持多个根节点:
function listToTree(list) {
var map = {}, node, tree= [], i;
for (i = 0; i < list.length; i ++) {
map[list[i].id] = list[i];
list[i].children = [];
}
for (i = 0; i < list.length; i += 1) {
node = list[i];
if (node.pid !== '-1') {
map[node.pid].children.push(node);
} else {
tree.push(node);
}
}
return tree;
}
listToTree(arr); //[{"id":1,"pid":"-1","children":[{"id":11,"pid":"1","children":[]},{"id":12,"pid":"1","children":[]}]}]
但是项目中有个需求,在后台没有返回给带层级信息level
的时候,需要用到层级信息,这样转换没法计算出层级,因此就需要用迭代的方法了,默认根节点层级为0,依次递增:
function listToTreeWithLevel(list, parent, level) {
var out = []
for (var node of list) {
if (node.pid == parent) {
node.level = level;
var children = listToTreeWithLevel(list, node.id, level + 1)
if (children.length) {
node.children = children
}
out.push(node)
}
}
return out
}
listToTreeWithLevel(arr, '-1', 0) //[{"id":1,"pid":"-1","children":[{"id":11,"pid":"1","children":[],"level":1},{"id":12,"pid":"1","children":[],"level":1}],"level":0}]
树形结构转换为扁平数组
这个其实就是数据结构中的广度优先遍历:
function treeToList(tree) {
var queen = [];
var out = [];
queen = queen.concat(tree);
while(queen.length) {
var first = queen.shift();
if (first.children) {
queen = queen.concat(first.children)
delete first['children'];
}
out.push(first);
}
return out;
}
var tree = [{"id":1,"pid":"-1","children":[{"id":11,"pid":"1","children":[]},{"id":12,"pid":"1","children":[]}]}];
treeToList(tree) //[{"id":1,"pid":"-1"},{"id":11,"pid":"1"},{"id":12,"pid":"1"}]
参考资料
- listtotree
- 找本数据结构看看dfs和bfs
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