Question

我编写了一个函数来检查我的树是否包含值。每个树/子树都有一个我的函数遍历的子数组。为什么我不能使用forEach来遍历树的孩子？

＆＃13;

function Tree (value) {
  this.value = value;
  this.children = [];
}

Tree.prototype.addChild = function (node) {
  this.children.push(node);
  return true;
};

Tree.prototype.contains = function (value) {
  if (this.value === value) return true;
  this.children.forEach(el => {
    return el.contains(value);
  });
  return false;
};

let myTree = new Tree(2);
let firstLevelChild = new Tree(7);
let secondLevelChild = new Tree(3);
myTree.addChild(firstLevelChild);
firstLevelChild.addChild(secondLevelChild);
console.log(JSON.stringify(myTree, null, 2));
console.log('2: ',myTree.contains(2))
console.log('7: ',myTree.contains(7))
console.log('3: ',myTree.contains(3))

＆＃13;

Answer 1

因为forEach并不关心其回调的返回值。它保持循环，无论如何，并且它本身没有（有意义的）返回值。

您希望some并返回其结果：

Tree.prototype.contains = function (value) {
  if (this.value === value) return true;
  return this.children.some(el => {
    return el.contains(value);
  });
};

甚至更紧凑（如果这是一个目标）：

Tree.prototype.contains = function (value) {
  return this.value === value || this.children.some(el => el.contains(value));
};

直播示例：

＆＃13;

function Tree (value) {
  this.value = value;
  this.children = [];
}

Tree.prototype.addChild = function (node) {
  this.children.push(node);
  return true;
};

Tree.prototype.contains = function (value) {
  return this.value === value || this.children.some(el => {
    return el.contains(value);
  });
};

let myTree = new Tree(2);
let firstLevelChild = new Tree(7);
let secondLevelChild = new Tree(3);
myTree.addChild(firstLevelChild);
firstLevelChild.addChild(secondLevelChild);
console.log('2: ',myTree.contains(2));
console.log('7: ',myTree.contains(7));
console.log('3: ',myTree.contains(3));

＆＃13;

some调用其回调，直到回调返回truthy值，然后停止。如果回调值返回真值，some将返回true;如果没有，some会返回false。

为什么我不能使用forEach来遍历一棵树？

1 个答案: