您的代码只是缺少一个循环来检查
child数组中节点的每个子节点。此递归函数将返回
label节点的属性，或者
undefined如果树中不存在标签，则返回该属性：
const search = (tree, target) => { if (tree.id === target) { return tree.label; } for (const child of tree.child) { const res = search(child, target); if (res) { return res; } }};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":[]}]}]};console.log(search(tree, 1));console.log(search(tree, 6));console.log(search(tree, 99));您还可以使用显式堆栈进行迭代，该堆栈更快，更凉爽并且不会导致堆栈溢出：
const search = (tree, target) => { const stack = [tree]; while (stack.length) { const curr = stack.pop(); if (curr.id === target) { return curr.label; } stack.push(...curr.child); }};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":[]}]}]};for (let i = 0; ++i < 12; console.log(search(tree, i)));
