广度优先搜索二进制搜索树javascript实现

Question

I have the following code that implements a BST tree in JavaScript.

function Node(value) {
  this.left = null;
  this.right = null;
  this.value = value;
}

function BinarySearchTree() {
  this.root = null;
  return;
}

BinarySearchTree.prototype.push = function(value) {

  if (!this.root) {
    this.root = new Node(value);
    return;
  }

  var currentRoot = this.root;
  var newNode = new Node(value);
  while (currentRoot) {
    if (value < currentRoot.value) {
      if (!currentRoot.left) {
        currentRoot.left = newNode;
        break;
      } else {
        currentRoot = currentRoot.left;
      }
    } else {

      if (!currentRoot.right) {
        currentRoot.right = newNode;
        break;
      } else {
        currentRoot = currentRoot.right;
      }

    }

  }

}

var a = new BinarySearchTree();
a.push(27);
a.push(14);
a.push(35);
a.push(10);
a.push(19);
a.push(31);
a.push(42);

I am trying to implement a function which can do a breadth first traversal of the tree. This is what I have tried so far.

console.log(a.root.value);
traverse(a.root);

//function to traverse 
function traverse(node) {

  currentNode = node;
  while (currentNode.left) {
    displayNodes(currentNode);
    parent = currentNode;
    currentNode = currentNode.left;
    displayNodes(currentNode);
    if(parent.right!=null){
    displayNodes(parent.right);
    }
  }
}

//function that displays the left and right node of a node 
function displayNodes(node) {


  if (node.left != null) {
    console.log(node.left.value);
  }
  if (node.right != null) {
    console.log(node.right.value);
  }
}

I am unable to implement a function that could scale with a large number of data. I am not sure if a recursive method to traverse would be better or using a while loop. How can I implement the function? I know that the function gives unexpected behavior? What correction should I make?

Answer 1

You currently traverse the path from the root node to the left-most leaf.

A simple non-recursive breadth-first traversal function invoking a callback on each traversed node could look as follows:

// Breadth-first traversal:
function traverse(node, cb) {
  var current = [node];
  while (current.length > 0) {
    var next = [];
    for (var node of current) {
      cb(node);
      if (node.left) next.push(node.left);
      if (node.right) next.push(node.right);
    }
    current = next;
  }
}

// Example:
traverse(root, function(node) {
  console.log(node.value);
});

It works by keeping an array of already discovered or traversed nodes current which initially contains just your root node. Now, you iteratively replace each node in that list with its children. In above function, the children are stored in a next array. At the end of each iteration, all nodes of the current level in current are replaced with all their children of the next deeper level in next . See also the first suggestion given by @DavidKnipe's answer .

A non-recursive approach has the advantage of not being subject to the call stack size limit. This theoretically allows you to handle larger data structures when the call stack size is limited .

Answer 2

If you're looking for a way to BFS using O(1) memory, I don't think there's a nice way to do it. (DFS is another matter though. Are you sure it has to be BFS?)

There are two ways I can see to do this. You could start with the array [this.root] , and write a function that iterates over an array of nodes and then returns an array of children of those nodes. Then call that function on the array of children, and keep going down the tree until you get an empty array.

If memory is an issue, there's another way to do it. Instead of remembering the array of nodes at a given level, you could just remember the depth, then redo the iteration each time. So you'd have a function which takes a natural number n and iterates over the tree, but without going deeper than n , and does whatever it is you're trying to do at the n th level only; then call this function for all values of n until there are no more nodes left.

That last one might sound very wasteful, but it might not be too bad if the last few levels of the tree contain most of the nodes. It depends on your dataset and computational capabilities.