Java从树中找到具有递归功能的节点 [英] Javascript find node from tree with recursive function

问题描述

我有这样的 JavaScript 树数据.

const tree = {
    children:[
        {id: 10, children: [{id: 34, children:[]}, {id: 35, children:[]}, {id: 36, children:[]}]},
        {id: 10, 
            children: [
                {id: 34, children:[
                        {id: 345, children:[]}
                    ]}, 
                {id: 35, children:[]}, 
                {id: 36, children:[]}
                ]
        },
        {id: 11, children: [{id: 30, children:[]}, {id: 33, children:[]}, {id: 3109, children:[]}]}
        ],
    id: 45
}

const getByID = (tree, id) => {
	let result = null
        if (id === tree.id) {
            return tree
        } else {
            if(tree.children){
                tree.children.forEach( node=> {
                    result = getByID(node, id)
                })
            }
            return result
        }
}

const find345 = getByID(tree, 345)
console.log(find345)

我试图从这棵树中按ID查找项目.我使用递归函数来迭代树及其子对象，但无法按我的期望找到该项目.

I was try to find item by its id from this tree. im using recursive function to iterate the tree and its children, but it wont find the item as i expected.

它总是返回null.预期返回 {id:345，子代:[]}

its always return null. expected to return {id: 345, children:[]}

推荐答案

您需要使用允许在查找时发生短路的方法来退出循环.

You need to exit the loop by using a method which allows a short circuit on find.

访问节点但已找到该节点的问题是用后来的错误结果替换结果.您需要使用找到的节点尽早退出.

The problem with visiting nodes but already found the node, is the replacement of the result with a later wrong result. You need to exit early with a found node.

Array#some 允许迭代并退出循环(如果返回了truty值).在这种情况下，结果是真实的.

Array#some allows to iterate and to exit the loop if a truty value is returned. In this case the result is truthy on find.

const tree = { children: [{ id: 10, children: [{ id: 34, children: [] }, { id: 35, children: [] }, { id: 36, children: [] }] }, { id: 10, children: [{ id: 34, children: [{ id: 345, children: [] }] }, { id: 35, children: [] }, { id: 36, children: [] }] }, { id: 11, children: [{ id: 30, children: [] }, { id: 33, children: [] }, { id: 3109, children: [] }] }], id: 45 };

const getByID = (tree, id) => {
	let result = null
        if (id === tree.id) {
            return tree
        } else {
            if(tree.children){
                tree.children.some(node => result = getByID(node, id));
                //            ^^^^                                      exit if found
                //                         ^^^^^^^^^^^^^^^^^^^^^^^^^^   return & assign
            }
            return result;
        }
}

const find345 = getByID(tree, 345)
console.log(find345)

短一点

var tree = { children: [{ id: 10, children: [{ id: 34, children: [] }, { id: 35, children: [] }, { id: 36, children: [] }] }, { id: 10, children: [{ id: 34, children: [{ id: 345, children: [] }] }, { id: 35, children: [] }, { id: 36, children: [] }] }, { id: 11, children: [{ id: 30, children: [] }, { id: 33, children: [] }, { id: 3109, children: [] }] }], id: 45 },
    getByID = (tree, id) => {
        var temp;
        return tree.id === id
            ? tree
            : (tree.children || []).some(o => temp = getByID(o, id)) && temp;
    };

console.log(getByID(tree, 345));

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