怎么实现node 子节点

观察以下JSON数据结构，设计Node类和⼯具⽅法 Node NodeUtils.getNode(List n odes, String path) ，通过该⽅法可以获取nodes中的Node.path与输⼊参数path相同的节点

[ { "path": "level0_0/", "value": "value", "children": [ { "path": "level0_0/level1_0/", "value": "value", "children": [ { "path": "level0_0/level1_0/level2_0", "value": "value", "children": [] },{ "path": "level0_0/level1_0/level2_1", "value": "value", "children": [] },{ "path": "level0_0/level1_0/level2_1", "value": "value", "children": [] 3 | } ] },{ "path": "level0_0/level1_1/", "value": "value", "children": [ { "path": "level0_0/level1_1/level2_0", "value": "value", "children": [] },{ "path": "level0_0/level1_1/level2_1", "value": "value", "children": [] },{ "path": "level0_0/level1_1/level2_1", "value": "value", "children": [] } ] } ] },{ "path": "level0_1/", "value": "avaluerg", "children": [] } ] 实现效果（伪代码） List nodes = JSON.parse(str); Node node = NodeUtils.getNode(nodes, "level0_0/level1_1/level2_1"); print(JSON.toString(node)); //运行结果 { "path": "level0_0/level1_1/level2_1", "value": "value", "children": [] } 4 |