Javascript-only Dom Tree Traversal - Dfs And Bfs?

Can anyone provide either code, pseudocode, or even provide good links to implementing DFS and BFS in plain JavaScript (No JQuery, or any helper libraries)? I've been trying to und

Solution 1:

Let's use the following HTML code as an example:

<divclass="a"><divclass="aa"><spanclass="aaa"></span><spanclass="aab"></span></div><divclass="ab"><spanclass="aba"></span><spanclass="abb"></span></div></div>

DFS will always go to the next level of nodes first, and only if there are no more un-traversed child nodes will it step to a next node on the current level.

A DFS would traverse the nodes of the example in the following order:

a, aa, aaa, aab, ab, aba, abb

BFS will always traverse all the nodes in the current level first and only after that will it go to the next level of nodes.

A BFS would traverse the nodes of the example in the following order:

a, aa, ab, aaa, aab, aba, abb

There isn't a definite answer which one of these you should use. Usually it depends on your needs.

Implementation details:

For a DFS people often use a stack.

Pseudo code:

stack my_stack;
list visited_nodes;
my_stack.push(starting_node);

while my_stack.length > 0
   current_node = my_stack.pop();

   if current_node == nullcontinue;
   if current_node in visited_nodes
      continue;
   visited_nodes.add(current_node);

   // visit node, get the class or whatever you needforeach child in current_node.children
      my_stack.push(child);

This code will go until there is any nodes in the stack. In each step we get the top node in the stack and if it's not null and if we haven't visited it before than we visit it and add all its children to the stack.

Queue is usually used for the BFS.

Pseudo code:

queue my_queue;
list visited_nodes;
my_queue.enqueue(starting_node);

while my_queue.length > 0
   current_node = my_queue.dequeue();

   if current_node == nullcontinue;
   if current_node in visited_nodes
      continue;
   visited_nodes.add(current_node);

   // visit node, get the class or whatever you needforeach child in current_node.children
      my_queue.enqueue(child);

This code will go until there is any nodes in the queue. In each step we get the first node in the queue and if it's not null and if we haven't visited it before than we visit it and add all its children to the queue.

Note that the main difference between the two algorithm is the data type we use.

Solution 2:

DFS:

function m(elem) {
    elem.childNodes.forEach(function(a) {
        m(a);
    });
    //do sth with elem
}
m(document.body);

This loops through all elements, and for each element through each child's and so on.

BFS:

var childs = [];

functionm(elem) {
    elem.childNodes.forEach(function(a) {
        childs.push(a);
    });
    b = childs;
    childs = [];
    b.forEach(function(a) {
        m(a);
    });
}
m(document.body);

This loops through the elements, pushes their children onto a stack, and starts again with each of them. As you can see, this consumes much more space (the child's array) which is not the best...

Solution 3:

For DFS you can use the TreeWalker or NodeIterator API and filter with NodeFilter.SHOW_ELEMENT.