javascript中树的广度优先遍历 [英] breadth-first traversal of a tree in javascript

问题描述

我正在努力学习数据结构，并为常规树上的回调的深度优先遍历/应用实现了以下代码:

I am trying to learn data structures well and implemented the following code for a depth-first traversal/application of a callback on a regular tree:

Tree.prototype.traverse = function (callback) {
  callback(this.value);

  if (!this.children) {
    return;
  }
  for (var i = 0; i < this.children.length; i++) {
    var child = this.children[i];
    child.traverse(callback);
  }
};

我该如何更改以使其广度优先?这是树类的样子:

How could I change this to make it breadth first instead? This is what the Tree Class looks like:

var Tree = function (value) {
  var newTree = {};

  newTree.value = value;
  newTree.children = [];
  extend(newTree, treeMethods);

  return newTree;
};

推荐答案

从根本上说，DFS 和 BFS 的区别在于，使用 DFS 时，您将当前节点的子节点推入堆栈，因此它们将被弹出和处理 在其他所有事情之前，而对于 BFS，您将孩子推到队列的末尾，因此它们将在其他所有事情之后 被弹出和处理.

Fundamentally, the difference between DFS and BFS is that with a DFS you push the children of the current node onto a stack, so they will be popped and processed before everything else, while for BFS you push the children onto the end of a queue, so they will be popped and processed after everything else.

DFS 易于递归实现，因为您可以使用调用堆栈作为堆栈.你不能用 BFS 做到这一点，因为你需要一个队列.为了使相似性清晰，让我们首先将您的 DFS 转换为迭代实现:

DFS is easy to implement recursively because you can use the call stack as the stack. You can't do that with BFS, because you need a queue. Just to make the similarity clear, lets convert your DFS to an iterative implementation first:

//DFS
Tree.prototype.traverse = function (callback) {
  var stack=[this];
  var n;

  while(stack.length>0) {

    n = stack.pop();
    callback(n.value);

    if (!n.children) {
      continue;
    }

    for (var i = n.children.length-1; i>=0; i--) {
       stack.push(n.children[i]);
    }
  }
};

现在是 BFS

//BFS
Tree.prototype.traverse = function (callback) {
  var queue=[this];
  var n;

  while(queue.length>0) {

    n = queue.shift();
    callback(n.value);

    if (!n.children) {
      continue;
    }

    for (var i = 0; i< n.children.length; i++) {
       queue.push(n.children[i]);
    }
  }
};

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