Equivalent subtree

Question

I have two trees. The tree Node is defined as

class Node{
  String treeId; 
  String type;       //Each node has type which has fixed value. For example, its color: RED, BLANK, GREEN
  Set<Node> children;
  String ref;        //The ref is a string and allowed value are "0", "1",..."10". The value is null if it is not leaf. 
};

For leaf, the children set is empty.

I am wondering whether there is some existing efficient work done how to identify equivalent substree for two given tree. The equivalent is defined as:

1) Both subtree leaves are setsets leaves of original tree. 
2) Both subtrees leaves have same ref value. 
3) for non-leaves node, the equivalent refers to both node have same type and equivalent children. 

Thanks. It would be better if there is some Java library addressing this problem.

---

The input should are two tree roots while output is the Node that is root of equivalent subtree. An the the tree's height is 100~ and it has more than 500 nodes.

---

What i did now is that I added a new field for class Node.

class Cache{
   Map<String, Set<String>> map = new LinkedHashMap<String, Set<Str>>();
}

The key of map is Node id while the value is a ref set this node of this nodeid can reach. The Cache initiated when Node is initialized.

During isEquivalent compare phase, check whether overlap exists between two root's ref set. Return false if none.

I think this can help reduce the number of comparison space.

Answer 1

I am not sure about 1) Both subtree leaves are leaves of original tree. requirement as it seems to conflict with how to identify equivalent substree for two given tree.. Otherwise following recursive method should be able to cover other two conditions. The haveSameOriginalTree(r1, r2) method may be implemented to satisfy the first condition that I couldn't understand. r1 and r2 are roots of two subtrees that need to be checked for equivalence.

bool areEquivalent(Node r1, Node r2)
{
    if(r1.children == null && r2.children == null)
    {
        return (haveSameOriginalTree(r1, r2) && (r1.ref == r2.ref));
    }
    if((r1.children == null && r2.children != null) || (r1.children != null && r2.children == null))
    {
        return false;
    }
    // if here then both must be non-leaf nodes
    if(r1.type != r2.type)
    {
        return false;
    }
    if(r1.children.getCount() != r2.children.getCount()) // not sure of correct syntax for Java Sets
    {
        return false;
    }
    for(int i=0; i<r1.children.getCount(); i++)
    {
        if(!areEquivalent(r1.children[i], r2.children[i])) // again please correct the syntax for Sets
        {
            return false;
        }
    }

    return true;
}

Let me know what you think.

Update

Here is an iterative version of the above solution. It uses stack data structure which is allocated on the heap rather than pushed on function's call stack, so not hugely different from recursive but still better. Also, since we only hold references to Nodes (rather than copying the whole object), this shouldn't be that much of an additional memory overhead if we are already loading the original tree into memory.

bool areEquivalent(Node r1, Node r2)
{
    Stack<Node> s1 = new Stack<Node>();
    Stack<Node> s2 = new Stack<Node>();
    Node n1, n2;

    s1.Push(r1);
    s2.Push(r2);
    while(true) // Need a better check
    {
        if(s1.getCount() != s2.getCount())
        {
            return false;
        }
        if(s1.getCount() == 0) // if both stacks are empty then we've traversed both trees without failure.
        {
            return true;
        }
        n1 = s1.Pop();
        n2 = s2.Pop();
        if(!areEquivalentNodes(n1, n2))
        {
            return false;
        }
        foreach(Node child in n1.children)
        {
            s1.Push(child);
        }
        foreach(Node child in n2.children)
        {
            s2.Push(child);
        }
    }
}

// only checks the two nodes are equivalent. their childrens' equivalence will be handled by other calls to this method.
bool areEquivalentNodes(Node n1, Node n2)
{
    if(n1.children.getCount() != n2.children.getCount())
    {
        return false;
    }
    if(n1.children.getCount() == 0) // if both are leaf nodes...
    {
        if(n1.ref != n2.ref)
        {
            return false;
        }
    }
    else // both are non-leaf
    {
        if(n1.type != n2.type)
        {
            return false;
        }
        // the condition that children of non-leaf nodes be equivalent will be covered by subsequent calls this method...
    }

    return true;
}

Please note that both solutions expect children of two equivalent nodes in the same order. If children are not ordered then we will need to sort them before calling above code.

Let me know if this is better.