Generating all possible topologies in a full binary tree having n nodes

I want to create all possible topologies of a full binary tree which must have exactly n+1 leaf nodes and n internal nodes.

I want to create it using recursion and tree must be simple binary tree not a binary search tree or BST.

Kindly suggest algorithm to achieve this task.

example: with 4 leaf nodes and 3 internal nodes.

     N                  N                  N
    / \                / \                 /\
   N   N               N  N               N  N
  /\   /\             /\                     /\
 N  N  N N           N  N                    N N
                    / \                        /\
                    N  N                       N N

PS: There is a simlar thread .It will be helpful If anybody can elaborate the tree generation algorithm suggested by coproc in this thread.

Thanks in advance.

Here is the code for generating all possibles topologies for given n. Total Number of nodes (internal + leaf nodes) in a full binary tree is odd.

If a tree is full binary tree, then it's left and right sub trees are also full binary trees i.e both left and right sub trees have odd number of nodes.

For a given n, We generate all combinations of full binary tree like this

First Iteration: 1 Node on left hand side, 1 root, n-2 on right hand side.
Second Iteration: 3 Nodes on left hand side, 1 root, n-4 on right hand side.
Third Iteration: 5 Nodes on left hand side, 1 root, n-6 on right hand side. .
.
.
Last Iteration: n-2 Nodes on left hand side, 1 root, 1 on right hand side

In each iteration, we find all possible left trees and right trees. If L full trees are possible on left hand side, R full trees are possible on right hand side - then total number of trees is L*R

public void createAllTopologies(int n){

    if(n%2 == 0) return;

    Map<Integer, List<Node>> topologies = new HashMap<Integer, List<Node>>();

    topologies.put(1, new ArrayList<Node>());
    topologies.get(1).add(new Node(1));

    for(int i=3;i<=n;i+=2){

        List<Node> list = new ArrayList<Node>();

        for(int j=1;j<i;j+=2){
            List<Node> left = topologies.get(j);
            List<Node> right = topologies.get(i-j-1);
            list.addAll(generateAllCombinations(left,right));
        }
        topologies.put(i, list);
    }

    List<Node> result = topologies.get(n);

    for(int i=0;i<result.size();i++){
        printTree(result.get(i),0);
        System.out.println("-----------------------------");
    }

}

private List<Node> generateAllCombinations(List<Node> left, List<Node> right) {

    List<Node> list = new ArrayList<Node>();
    for(int i=0;i<left.size();i++){
        for(int j=0;j<right.size();j++){
            Node nNode = new Node(1);
            nNode.left = left.get(i).clone();
            nNode.right = right.get(j).clone();
            list.add(nNode);
        }
    }
    return list;
}

/* This prints tree from left to right fashion i.e 
       root at left, 
       leftNode at bottom, 
       rightNode at top 
*/

protected void printTree(Node nNode,int pos){
        if (nNode==null) {
            for(int i=0;i<pos;i++) System.out.print("\t");
            System.out.println("*");
            return;
        }
        printTree(nNode.right,pos+1);
        for(int i=0;i<pos;i++) System.out.print("\t");
        System.out.println(nNode.data);
        printTree(nNode.left,pos+1);

}

Please refer to complete code here - http://ideone.com/Wz2Jhm

来源：https://stackoverflow.com/questions/18654085/generating-all-possible-topologies-in-a-full-binary-tree-having-n-nodes

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binary-tree