JavaScript - 递归函数中的错误是什么？

Question

我尝试获取查询 ID 的完整路径。例如，getPath(6, rawData) 应该返回 [1,2,6]。但是，它返回 [6,6,6]。当我在代码中尝试 console.log 时，看起来自调用函数内部节点的值将覆盖外部节点的值。我无法弄清楚递归函数有什么问题。或者有人可以教我实现相同目标的新方法吗？

代码如下:

const getPath = (id, list) => {
  if (!list || !Array.isArray(list)) {
    return false
  }

  for (node of list) {
    if (node.id === id) {
      return [node.id]
    }

    let res = getPath(id, node.children)

    if (res) {
      return [node.id, ...res]
    }
  }

  return false
}

const rawData = [{
    id: 1,
    parent: null,
    children: [{
      id: 2,
      parent: 1,
      children: [{
        id: 4,
        parent: 2,
        children: null
      }, {
        id: 5,
        parent: 2,
        children: null
      }, {
        id: 6,
        parent: 2,
        children: null
      }]
    }]
  },
  {
    id: 3,
    parent: null,
    children: [{
        id: 7,
        parent: 3,
        children: [{
          id: 9,
          parent: 7,
        }]
      },
      {
        id: 8,
        parent: 3,
      }
    ]
  }
]

console.log(getPath(2, rawData))

Answer 1

您需要声明变量 node 以便它是函数的局部变量。

const getPath = (id, list) => {
  if (!list || !Array.isArray(list)) {
    return false
  }

  for (let node of list) {
    if (node.id === id) {
      return [node.id]
    }

    let res = getPath(id, node.children)

    if (res) {
      return [node.id, ...res]
    }
  }

  return false
}

const rawData = [{
    id: 1,
    parent: null,
    children: [{
      id: 2,
      parent: 1,
      children: [{
        id: 4,
        parent: 2,
        children: null
      }, {
        id: 5,
        parent: 2,
        children: null
      }, {
        id: 6,
        parent: 2,
        children: null
      }]
    }]
  },
  {
    id: 3,
    parent: null,
    children: [{
        id: 7,
        parent: 3,
        children: [{
          id: 9,
          parent: 7,
        }]
      },
      {
        id: 8,
        parent: 3,
      }
    ]
  }
]

console.log(getPath(2, rawData))