扁平数组和树形结构的相互转换

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"}]