Your code is just missing a loop to inspect each child of a node in the child
array. This recursive function will return the label
property of a node or undefined
if label not present in tree:
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));
You can also do it iteratively with an explicit stack which won’t cause a stack overflow (but note that the shorthand stack.push(...curr.child);
can overflow the argument size for some JS engines due to the spread syntax, so use an explicit loop or concat
for massive child arrays):
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; ++i < 12; console.log(search(tree, i)));
A somewhat more generic design would return the node itself and let the caller access the .label
property if they want to, or use the object in some other manner.
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.