JS广度优先遍历

自己用JS实现了 广度优先遍历

 第一种用了数组的高阶函数，看起来有些复杂。然后思索着从可读性上优化了一下，孰优孰劣以后分析。

var list = [{
    id: "ab",
    children: [{
        id: "ab1",
        children: [{
            id: "ab11",
            children: []
        }]
    },{
        id: "ab2",
        children: []
    }, {
        id: "cd",
        children: [{
            id: "aa",
            children: []
        }, {
            id: "ef",
            children: []
        }]
    }]
},{
    id:'cc',
    children:[]
},{
    id:'dd',
    children:[]
}]

// path[] 1.将list 数组合并到path， 2.从头部开始删除，看是否有子节点，有就合并到path[] 3.重复2，直到path 为空

function breadthFirstSearch(list)
{
    let tree=[];
    let path = list.reduce((acc,cur)=>acc=[].concat(acc,cur));

    while(path.length>0){
        let node = path.shift();
        tree.push(node.id);
        let target = node.children;
        // console.log(target.length,node.id,target)
        if(target.length !== 0){
            path=[].concat(path, ...target.length==1?target:target.reduce((acc,cur)=>acc=[].concat(acc,cur)));
        }
    }
    return tree;
}

console.log(breadthFirstSearch(list))

// [
//     'ab',   'cc',
//     'dd',   'ab1',
//     'ab2',  'cd',
//     'ab11', 'aa',
//     'ef'
//   ]
// 非递归： 1.创建新的数组nodeList, 并赋值list 2.从头遍历nodeList的子节点，并合并到nodeList
function breadthFirstSearch1(list){
    let index=0;
    let tree=[];
    let nodeList = [...list];
    while(index<nodeList.length){
        const node=nodeList[index++];
        tree.push(node.id);
        if(node.children){
            for(let k in node.children){
                nodeList.push(node.children[k]);
            }
        }
    }
    return tree;
    // return nodeList;
}
console.log(breadthFirstSearch1(list))