Javascript树类2

Question

这个问题是代码这里的第二个版本。

我在写一个普通的树类。具体来说，每个节点都应该有oen父节点、一定数量的子节点，并保存一个值。

我正在寻找关于如何使这段代码更加地道的一般性建议。我还在寻找关于traverse函数的想法，任何应该出现在普通树类中但不存在的函数，以及简化现有函数的任何方法。

// Code to review
class Node {
	constructor(value = 0, children = []) {
		this.value = value;
		this.children = children;
	}

	traverse({ preorder, postorder, levelorder }) {
		if (levelorder) {
			let nodes = [this];
			while (nodes.length > 0) {
				const current = nodes.shift();
				levelorder(current);
				nodes = nodes.concat(current.children);
			}
		}
		else {
			if (preorder) preorder(this);
			this.children.forEach(n => n.traverse({ preorder, postorder }));
			if (postorder) postorder(this);
		}
		return this;
	}

	clone() {
		const copy = n => Object.assign(new Node(), n);
		let that = copy(this);
		that.traverse({ preorder: n => n.children = n.children.map(copy) });
		return that;
	}

	map(callback) {
		let that = this.clone();
		that.traverse({ levelorder: n => n.value = callback(n.value) });
		return that;
	}

	reduce(callback, initial) {
		let a = initial;
		this.traverse({	levelorder: n => a = (n === this && initial === undefined)? n.value: callback(a, n.value) });
		return a;
	}

	filter(callback) {
		let that = this.clone();
		that.traverse({ levelorder: n => n.children = n.children.filter(m => callback(m.value)) });
		return that;
	}

	every(callback) {
		return Boolean(this.reduce((a, b) => a && callback(b), true));
	}

	some(callback) {
		return Boolean(this.reduce((a, b) => a || callback(b), false));
	}

	find(callback) {
		return this.reduce((a, b) => (callback(b)? a.push(b): null, a), []);
	}

	includes(value) {
		return this.some(a => a === value);
	}
}

// Testing code
let tree = new Node(0);
let a = new Node(1);
let b = new Node(2);
let c = new Node(3);
let d = new Node(4);
let e = new Node(5);
let f = new Node(6);
tree.children = [a, b, c];
a.children = [d];
d.children = [e, f];

console.log("\nmap", tree.map(a => a + "2"));
console.log("\nreduce", tree.reduce((a, b) => String(a) + String(b)));
console.log("\nfilter", tree.filter(a => a < 3));

console.log("\nevery", tree.every(a => a < 3));
console.log("\nsome", tree.some(a => a < 3))

console.log("\nfind", tree.find(a => a > 2));
console.log("\nincludes", tree.includes(3));

Answer 1

评论

编写得很好的API，但我不满意这个：

导线({预序，后序，水平序})

preorder和postorder可以结合在一起，但它们总是与levelorder相互排斥。

遍历({预序，后序，水准}) {如果(水平顺序){ //水准顺序。}{ //预定条件和/或后继条件。}返回此；}

相反，可以使用两种不同的方法，每种方法都执行自己熟悉的遍历类型。

外勤部

 traverseDepthFirst({ preorder, postorder })

BFS

 traverseBreadthFirst({ levelorder })

一些一般性的想法：

• 你应该允许循环图吗？
• 你想要遍历祖先吗？