前天面試遇到一個多叉樹面試的題目，在這里分享記錄一下。

題目：一個樹形的數(shù)據(jù)(如下數(shù)據(jù))，面試官給你一個id，然后拿到對應(yīng)的name?

數(shù)據(jù)結(jié)構(gòu)大概是這個樣子

var cityData = [
      {
        id: 1,
        name: "廣東省",
        children: [
          {
            id: 11,
            name: "深圳",
            children: [
              {
                id: 111,
                name: "寶安",
                children: [
                  {
                    id: 1111,
                    name: "西鄉(xiāng)",
                    children:[
                      {
                        id: 11111,
                        name: "坪洲",
                        children:[]
                      },
                      {
                        id: 11112,
                        name: "靈芝",
                        children:[]
                      }
                    ]
                  },
                  {
                    id: 1112,
                    name: "南山",
                    children:[
                      {
                        id: 11121,
                        name: "科技園",
                        children:[]
                      }
                    ]
                  }
                ]
              },
              {
                id: 112,
                name: "福田",
                children: []
              }
            ]
          },
          {
            id: 12,
            name: "廣州",
            children: [
              {
                id: 122,
                name: "白云區(qū)",
                children: [
                  {
                    id: 1222,
                    name: "白云區(qū)",
                    children: []
                  }
                ]
              },
              {
                id: 122,
                name: "珠海區(qū)",
                children: []
              }
            ]
          }
        ]
      },
      {
        id: 2,
        name: "湖南省",
        children: []
      }
    ];

比如說 我要id是11112的name返回是靈芝，請問你有幾種解法？？

這題目讓人看到這不就是考用遞歸的方法嘛，代碼如下

let result = ""

// 遞歸實現(xiàn)
const recursion = (cityData, id) => {
  // cityData數(shù)據(jù)為空的時候直接返回
  if (!cityData || !cityData.length) return;
  // 常規(guī)循環(huán)cityData
  for (let i = 0, len = cityData.length; i < len; i++) {
    const childs = cityData[i].children;
    
    // 如果匹配到id的話，就是我們要的結(jié)果
    if (cityData[i].id === id) {
      result = cityData[i].name
    }
    // 如果還有子節(jié)點，執(zhí)行遞歸
    if(childs && childs.length > 0){
      recursion(childs, id);
    }
  }
  return result
};

const r = recursion(cityData, 11112);
console.log(r) // 靈芝

oyes~~~完成了么？？面試官可能不滿意哦，下面還有幾種解法

let result = ""

const range = (cityData, id) => {
  if (!cityData || !cityData.length) return;
  // 定義一個數(shù)據(jù)棧
  let stack = [];

  let item = null;

  //先將第一層節(jié)點放入棧
  for (var i = 0, len = cityData.length; i < len; i++) {
    stack.push(cityData[i]);
  }

  while (stack.length) {
    // 將數(shù)據(jù)棧的第一個取出來
    item = stack.shift();
    // 如果符合就賦值給result
    if (item.id === id) {
      result = item.name
    }
    //如果該節(jié)點有子節(jié)點，繼續(xù)添加進入棧底
    if (item.children && item.children.length) {
      stack = stack.concat(item.children);
    }
  }
  return result
};

let r1 = range(cityData, 11112);

console.log(r1) // 靈芝

let result = ""

const deep = (cityData, id) => {
  // 沒有數(shù)據(jù)直接返回
  if (!cityData || !cityData.length) return;
  // 先定義一個數(shù)據(jù)棧
  let stack = []
  let item = null

  //先將第一層節(jié)點放入數(shù)據(jù)棧
  for (var i = 0, len = cityData.length; i < len; i++) {
    stack.push(cityData[i])
  }
  // 循環(huán)
  while (stack.length) {
    item = stack.shift()
    if (item.id === id) {
      result = item.name
    }
    //如果該節(jié)點有子節(jié)點，繼續(xù)添加進入棧頂
    if (item.children && item.children.length) {
      // 注意這里調(diào)換了順序
      stack = item.children.concat(stack);
    }
  }
  return result
};

let r3 = deep(cityData, 11112)
console.log(r3) // 靈芝

const regular = (cityData, id) => {
  // 沒有數(shù)據(jù)直接返回
  if (!cityData || !cityData.length) return;
  // 數(shù)據(jù)轉(zhuǎn)成字符串
  let cityStr = JSON.stringify(cityData)
  // 定義正則
  let reg = new RegExp(`"id":${id},"name":"([^x00-xff]+)",`)
  // 取到正則的子字符串并返回
  return (cityStr.match(reg))[1]
}

let r4 = regular(cityData, 11112);

console.log(r4) // 靈芝

這里列舉了4種方法，應(yīng)該還有很多種方法，大佬們有的話可以留言給我，先謝謝啦~~

安利一波博客~~~https://github.com/naihe138/naice-blog