Javascript javacscript中的筛选器对象

我有一个包含子对象的对象。我想过滤它们。我已经写了一个过滤代码，但仍停留在如何创建具有相同关系的新对象之间

老照片-

预期结果-

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这很有挑战性！有一种方法可以删除未通过过滤器的节点。如果不想，只需克隆树结构即可

首先，让我们创建一个根元素，以便在第一个元素与筛选条件不匹配时不会拆分树：

var tree = {"shape":"root","health":"ultraviolet","children":obj};
// obj is your structure in your question

现在removeNode函数将是

function removeNode(victim,parent) {
  // remove the victim from parent's children
  var target = parent.children.indexOf(victim);
  if(target>-1) parent.children.splice(target,1);
  // add victim's children to parent's children
  if(victim.children) parent.children = parent.children.concat(victim.children);
  // don't do this if you need a clone
  delete(victim);
}

function filter(root,shape,health) {
  if(root.children) root.children.forEach(function(o) {
    filter(o,shape,health);
    if(!(o.shape==shape && o.health==health) && o.shape!="root") {
      removeNode(o,root);
      // possible optimization: you could skip nodes which passed already
      filter(root,shape,health);
    }
  });
}

过滤功能将是

function removeNode(victim,parent) {
  // remove the victim from parent's children
  var target = parent.children.indexOf(victim);
  if(target>-1) parent.children.splice(target,1);
  // add victim's children to parent's children
  if(victim.children) parent.children = parent.children.concat(victim.children);
  // don't do this if you need a clone
  delete(victim);
}

function filter(root,shape,health) {
  if(root.children) root.children.forEach(function(o) {
    filter(o,shape,health);
    if(!(o.shape==shape && o.health==health) && o.shape!="root") {
      removeNode(o,root);
      // possible optimization: you could skip nodes which passed already
      filter(root,shape,health);
    }
  });
}

所以你可以

filter(tree,"circle","red");

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以获得您想要的结果。

当您通过一些示例输入明确问题时，您将得到更多的回答，预期的输出和您面临的实际问题。我会看一看二叉搜索树删除，这是一个非常类似的概念：我还建议寻找一种从树中删除元素的方法，而不是构建一个新的整体树。jan对于您的答案，我已经等待了很长时间了，我会在晚上核对答案，然后再给你回复