检测树中的节点是否有子节点或已访问子节点

Question

I am struggling with the logic behind detecting nodes in my tree exists or have been visited.

I have a tree with nodes (a node has a left and right child node).

I want to check 2 things on a node: If there are no child nodes If there are child(s) nodes I want to check if they have been visited.

I currently have a large condition which I hate the look of. Is there a way I can simplify it?

public boolean finished(){
      return right == null && left == null || ((right != null && right.visited && (left != null && left.visited))
}

I want finished() to be true: If a right and left node dont exist If a right exists and has been visited If a left exists and has been visited

I think I also need an OR so if a right exists AND visited AND left is null

I'm a bit confused :S

Answer 1

I think this may be what you are after

if( (right == null || right.visited) && ( left == null || left.visited) )

So for each child node if it is either empty or has been visited then you are finished.

This covers the cases

Right == null; Left == null; 
Right == null; Left.visited; 
Right.visited; Left == null; 
Right.visited; Left.visited;

which I think given the description is the cases you wanted to check for.

Answer 2

The simplest approach to "visited" is to put a "visited" flag in your nodes. Reset all flags to start, then do your tree traversal and set the flags as you visit them.

To avoid having to reset the "visited" flags after every use, make the "flag" an integer, and keep a global counter which is incremented each time you do a scan where you need the visited flag:

boolean haveVisited() {
   return visitedFlag == globalVisitedCounter;
}

void markVisited() {
   visitedFlag = globalVisitedCounter;
}

You do need to, in theory, have logic to scan all the nodes in the tree and reset them when the global counter wraps, but in practice that's usually not needed.