blocks|key|4064855|text|您的代码缺少一个用于检查child数组中每个节点的子节点的循环。如果树中不存在标签，此递归函数将返回节点或undefined的label属性：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4064856|​|4064857|const+search+=+(tree,+target)+=>+{
++if+(tree.id+===+target)+{
++++return+tree.label;
++}
++
++for+(const+child+of+tree.child)+{
++++const+found+=+search(child,+target);
++++
++++if+(found)+{
++++++return+found;
++++}
++}
};

const+tree+=+{"id":1,"label":"A","child":[{"id":2,"label":"B","child":[{"id":5,"label":"E","child":[]},{"id":6,"label":"F","child":[]},{"id":7,"label":"G","child":[]}]},{"id":3,"label":"C","child":[]},{"id":4,"label":"D","child":[{"id":8,"label":"H","child":[]},{"id":9,"label":"I","child":[]}]}]};

console.log(search(tree,+1));
console.log(search(tree,+6));
console.log(search(tree,+99));|code-block|syntax|javascript|4064858|4064859|您也可以使用显式堆栈迭代地执行此操作，这不会导致堆栈溢出(但请注意，由于扩展语法，速记stack.push(...curr.child);可能会溢出某些JS引擎的参数大小，因此对于大量子数组，请使用显式循环或concat+)：|4064860|4064861|const+search+=+(tree,+target)+=>+{
++for+(const+stack+=+[tree];+stack.length;)+{
++++const+curr+=+stack.pop();
++++
++++if+(curr.id+===+target)+{
++++++return+curr.label;
++++}

++++stack.push(...curr.child);
++}
};

const+tree+=+{"id":1,"label":"A","child":[{"id":2,"label":"B","child":[{"id":5,"label":"E","child":[]},{"id":6,"label":"F","child":[]},{"id":7,"label":"G","child":[]}]},{"id":3,"label":"C","child":[]},{"id":4,"label":"D","child":[{"id":8,"label":"H","child":[]},{"id":9,"label":"I","child":[]}]}]};

for+(let+i+=+0;+%2B%2Bi+<+12;+console.log(search(tree,+i)));|4064862|4064863|一种更通用的设计是返回节点本身，并允许调用者访问.label属性，或者以其他方式使用对象。|4064864|请注意，JSON纯粹是序列化(stringified，原始)数据的字符串格式。一旦将JSON反序列化为JavaScript对象结构，它就不再是JSON了。|4064865|entityMap^0|C|5|1H|9|1R|5|0|0|0|0|17|Q|2W|6|0|0|0|0|O|6|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|17|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|P|3|G|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1F|8|@]|D|@]|E|$K|L]]|$1|S|3|G|5|6|7|1G|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1H|8|@$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|V|3|W|5|6|7|1K|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|Y|$]]

<p>Your code is just missing a loop to inspect each child of a node in the <code>child</code> array. This recursive function will return the <code>label</code> property of a node or <code>undefined</code> if label not present in tree:</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const search = (tree, target) =&gt; {
  if (tree.id === target) {
    return tree.label;
  }
  
  for (const child of tree.child) {
    const found = search(child, target);
    
    if (found) {
      return found;
    }
  }
};

const tree = {"id":1,"label":"A","child":[{"id":2,"label":"B","child":[{"id":5,"label":"E","child":[]},{"id":6,"label":"F","child":[]},{"id":7,"label":"G","child":[]}]},{"id":3,"label":"C","child":[]},{"id":4,"label":"D","child":[{"id":8,"label":"H","child":[]},{"id":9,"label":"I","child":[]}]}]};

console.log(search(tree, 1));
console.log(search(tree, 6));
console.log(search(tree, 99));</code></pre>
</div>
</div>
</p>
<p>You can also do it iteratively with an explicit stack which won't cause a stack overflow (but note that the shorthand <code>stack.push(...curr.child);</code> can overflow the argument size for some JS engines due to the spread syntax, so use an explicit loop or <code>concat</code> for massive child arrays):</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const search = (tree, target) =&gt; {
  for (const stack = [tree]; stack.length;) {
    const curr = stack.pop();
    
    if (curr.id === target) {
      return curr.label;
    }

    stack.push(...curr.child);
  }
};

const tree = {"id":1,"label":"A","child":[{"id":2,"label":"B","child":[{"id":5,"label":"E","child":[]},{"id":6,"label":"F","child":[]},{"id":7,"label":"G","child":[]}]},{"id":3,"label":"C","child":[]},{"id":4,"label":"D","child":[{"id":8,"label":"H","child":[]},{"id":9,"label":"I","child":[]}]}]};

for (let i = 0; ++i &lt; 12; console.log(search(tree, i)));</code></pre>
</div>
</div>
</p>
<p>A somewhat more generic design would return the node itself and let the caller access the <code>.label</code> property if they want to, or use the object in some other manner.</p>
<p>Note that JSON is purely a string format for serialized (stringified, raw) data. Once you've deserialized JSON into a JavaScript object structure, as is here, it's no longer JSON.</p>


blocks|key|4064896|text|可以使用第三个参数递归编写scan，该参数对要扫描的节点队列进行建模|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4064897|const+scan+=+(id,+tree+=+{},+queue+=+[+tree+])+=>
++//+if+id+matches+node+id,+return+node+label
++id+===+tree.id
++++?+tree.label

++//+base+case:+queue+is+empty
++//+id+was+not+found,+return+false
++:+queue.length+===+0
++++?+false

++//+inductive+case:+at+least+one+node
++//+recur+on+next+tree+node,+append+node+children+to+queue
++:+scan+(id,+queue[0],+queue.slice(1).concat(queue[0].child))|code-block|syntax|javascript|4064898|因为JavaScript支持默认参数，所以scan的调用点是不变的|4064899|console.log
++(+scan+(1,+tree)++//+"A"
++,+scan+(3,+tree)++//+"C"
++,+scan+(9,+tree)++//+"I"
++,+scan+(99,+tree)+//+false
++)|4064900|在下面的浏览器中验证它是否正常工作|4064901|​|4064902|const+scan+=+(id,+tree+=+{},+queue+=+[+tree+])+=>
++id+===+tree.id
++++?+tree.label
++:+queue.length+===+0
++++?+false
++:+scan+(id,+queue[0],+queue.slice(1).concat(queue[0].child))

const+tree+=
++{+id:+1
++,+label:+"A"
++,+child:
++++++[+{+id:+2
++++++++,+label:+"B"
++++++++,+child:
++++++++++++[+{+id:+5
++++++++++++++,+label:+"E"
++++++++++++++,+child:+[]
++++++++++++++}
++++++++++++,+{+id:+6
++++++++++++++,+label:+"F"
++++++++++++++,+child:+[]
++++++++++++++}
++++++++++++,+{+id:+7
++++++++++++++,+label:+"G"
++++++++++++++,+child:+[]
++++++++++++++}
++++++++++++]
++++++++}
++++++,+{+id:+3
++++++++,+label:+"C"
++++++++,+child:+[]
++++++++}
++++++,+{+id:+4
++++++++,+label:+"D"
++++++++,+child:
++++++++++++[+{+id:+8
++++++++++++++,+label:+"H"
++++++++++++++,+child:+[]
++++++++++++++}
++++++++++++,+{+id:+9
++++++++++++++,+label:+"I"
++++++++++++++,+child:+[]
++++++++++++++}
++++++++++++]
++++++++}
++++++]
++}

console.log
++(+scan+(1,+tree)++//+"A"
++,+scan+(3,+tree)++//+"C"
++,+scan+(9,+tree)++//+"I"
++,+scan+(99,+tree)+//+false
++)|4064903|4064904|相关recursive+search+using+higher-order+functions|4064905|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/50538352/633183^0|D|4|0|0|L|4|0|0|0|0|0|0|2|19|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1D|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1E|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|U|3|R|5|6|7|1G|8|@]|D|@]|E|$]]|$1|V|3|W|5|6|7|1H|8|@]|D|@$9|1I|A|1J|1|1K]]|E|$]]|$1|X|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]]]

<p><code>scan</code> can be written recursively using a third parameter that models a queue of nodes to scan</p>

<pre><code>const scan = (id, tree = {}, queue = [ tree ]) =&gt;
  // if id matches node id, return node label
  id === tree.id
    ? tree.label

  // base case: queue is empty
  // id was not found, return false
  : queue.length === 0
    ? false

  // inductive case: at least one node
  // recur on next tree node, append node children to queue
  : scan (id, queue[0], queue.slice(1).concat(queue[0].child))
</code></pre>

<p>Becauase JavaScript supports default arguments, the call site for <code>scan</code> is unaltered</p>

<pre><code>console.log
  ( scan (1, tree)  // "A"
  , scan (3, tree)  // "C"
  , scan (9, tree)  // "I"
  , scan (99, tree) // false
  )
</code></pre>

<p>Verify it works in your browser below</p>

<p><div class="snippet" data-lang="js" data-hide="true" data-console="true" data-babel="false">
<div class="snippet-code snippet-currently-hidden">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const scan = (id, tree = {}, queue = [ tree ]) =&gt;
  id === tree.id
    ? tree.label
  : queue.length === 0
    ? false
  : scan (id, queue[0], queue.slice(1).concat(queue[0].child))

const tree =
  { id: 1
  , label: "A"
  , child:
      [ { id: 2
        , label: "B"
        , child:
            [ { id: 5
              , label: "E"
              , child: []
              }
            , { id: 6
              , label: "F"
              , child: []
              }
            , { id: 7
              , label: "G"
              , child: []
              }
            ]
        }
      , { id: 3
        , label: "C"
        , child: []
        }
      , { id: 4
        , label: "D"
        , child:
            [ { id: 8
              , label: "H"
              , child: []
              }
            , { id: 9
              , label: "I"
              , child: []
              }
            ]
        }
      ]
  }

console.log
  ( scan (1, tree)  // "A"
  , scan (3, tree)  // "C"
  , scan (9, tree)  // "I"
  , scan (99, tree) // false
  )</code></pre>
</div>
</div>
</p>

<p>Related <a href="https://stackoverflow.com/a/50538352/633183">recursive search using higher-order functions</a></p>


<p>I'm trying to figure out how to search for a node in this JSON object recursively. I have tried something but cannot get it:</p>

<pre><code>var tree = {
    "id": 1,
    "label": "A",
    "child": [
        {
            "id": 2,
            "label": "B",
            "child": [
                {
                    "id": 5,
                    "label": "E",
                    "child": []
                },
                {
                    "id": 6,
                    "label": "F",
                    "child": []
                },
                {
                    "id": 7,
                    "label": "G",
                    "child": []
                }
            ]
        },
        {
            "id": 3,
            "label": "C",
            "child": []
        },
        {
            "id": 4,
            "label": "D",
            "child": [
                {
                    "id": 8,
                    "label": "H",
                    "child": []
                },
                {
                    "id": 9,
                    "label": "I",
                    "child": []
                }
            ]
        }
    ]
};
</code></pre>

<p>Here is my non-working solution, which is probably because the first node is just a value while children are in arrays:</p>

<pre><code>function scan(id, tree) {
    if(tree.id == id) {
        return tree.label;
    }

    if(tree.child == 0) {
        return
    }

    return scan(tree.child);
};
</code></pre>


Recursive tree search in a nested object structure in JavaScript

JSON

我正在尝试弄清楚如何递归地搜索这个JSON对象中的节点。我已经尝试了一些东西，但无法获得它：var tree = {    "id": 1,    "label": "A",    "child": [        {            "id": 2,            "label": "B",     ...

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