Isomorphic nodes - Two trees where one of the trees can be obtained from the other by a series of flips, i.e. by swapping left and right children of a number of nodes
This project goes over the implementation of both recursive and iterative algorithms for the following.
- Check if two nodes are isomorphic with two methods
- Printing out a binary tree recursively - just because I thought it would be useful to have
There are a few different classes in this program that might be worth looking at later (Isomorphic, IsomorphicNodes, Node, and Printer).
You run the program, now what? The console should look something like the snippet below with minor differences in the times.
The driver class IsomorphicNodes creates and prints out 4 different nodes, the first, second and fourth nodes are isomorphic. Then the nodes are printed out with the Printer class. All time measurements are made using System.nanoTime(); for precision.
- Before I began timing the code, I hypothesized that the recursive algorithm would be slower than the iterative implementation.
Before we run the tests, in many computer science classes it's taught that you should only use recursion as a last resort. Iteration is usually the first way to go. Use loops for everything they said.
Run the program and behold the magic...
1) Printing tree1
1
/ \
/ \
/ \
/ \
2 3
/ \ /
/ \ /
4 5 6
/ \
7 8
Took 6460013 ns to print
2) Printing tree2
1
/ \
/ \
/ \
/ \
3 2
\ / \
\ / \
6 4 5
/ \
8 7
Took 1839256 ns to print
3) Printing tree3
1
/ \
/ \
/ \
/ \
2 3
\ / \
\ / \
4 6 5
/ \
8 7
Took 2669751 ns to print
4) Printing tree4
1
/ \
/ \
/ \
/ \
2 3
/ \ /
/ \ /
4 5 6
/ \
8 7
Took 1566381 ns to print
========================== Isomorphic Test ==========================
1) Comparing tree1 & tree2...
Recursive result IS isomorphic
Iterative result IS isomorphic
+----------------+----------------+
| | winner |
+----------------+----------------+
| recursive time | iterative time |
|----------------|----------------|
| 599080 ns | 80578 ns |
+----------------+----------------+
2) Comparing tree1 & tree3...
Recursive result IS NOT isomorphic
Iterative result IS NOT isomorphic
+----------------+----------------+
| winner | |
+----------------+----------------+
| recursive time | iterative time |
|----------------|----------------|
| 1497 ns | 11582 ns |
+----------------+----------------+
3) Comparing tree2 & tree3...
Recursive result IS NOT isomorphic
Iterative result IS NOT isomorphic
+----------------+----------------+
| winner | |
+----------------+----------------+
| recursive time | iterative time |
|----------------|----------------|
| 1183 ns | 11976 ns |
+----------------+----------------+
4) Comparing tree4 & tree1...
Recursive result IS isomorphic
Iterative result IS isomorphic
+----------------+----------------+
| winner | |
+----------------+----------------+
| recursive time | iterative time |
|----------------|----------------|
| 2321 ns | 51718 ns |
+----------------+----------------+
5) Comparing tree4 & tree2...
Recursive result IS isomorphic
Iterative result IS isomorphic
+----------------+----------------+
| winner | |
+----------------+----------------+
| recursive time | iterative time |
|----------------|----------------|
| 2051 ns | 39800 ns |
+----------------+----------------+
Test results
- The iterative method was faster by 518502 ns
- The recursive method was faster.
- The recursive method was faster.
- The recursive method was faster.
- The recursive method was faster.
Well, let's look at why I was wrong...
- Make two stacks for the opened nodes
- Push the first tree into the first stack then the second tree into the second stack
- Create 2 temporary node variables to store the value of nodes that are popped from the stack
- While the first stack and the second stack have nodes
- Pop a node from the first stack and set it to the first tmp node variable
- Pop a node from the second stack and set it to the second tmp node variable
- If the two nodes are NULL check the other nodes so continue the while loop
- If exactly one of the nodes is NULL then return false, there's no hope for these trees
- If the data from the two nodes isn't equal return false
- If the nodes are Symmetric
- Put the child nodes of the first and second node into the stack in symmetric order
- If the nodes are Flipped
- Put the child nodes of the first and second node into the stack in flipped order
- If you get this far, that means that the two nodes are isomorphic return true and finish
- If both roots are null then return true
- If one of the roots is null then return false
- If the data from the two nodes isn't equal then return false
- Return the value of a recursive check to see if the child nodes are flipped or symmetric values
Obviously, there are more steps involved with the iterative method so this adds more time to the code. The recursive method has 4 simple steps, less code and so it went straight to the point.
This is the code that runs when the recursive method is called to check for isomorphic nodes
/**
* recursively checks if the 2 nodes are isomorphic recursively
*
* @param first the first node/tree to compare
* @param second the second node/tree to compare
*/
static boolean recursive(Node first, Node second) {
// Both roots are NULL, trees isomorphic by definition
if (first == null && second == null) return true;
// Exactly one of the first and second is NULL, trees not isomorphic
if (first == null || second == null) return false;
// if the two are not equal, trees are not isomorphic
if (first.data != second.data) return false;
// There are two possible cases for first and second to be isomorphic
// Case 1: The subtrees rooted at these nodes have NOT been "Flipped".
// Both of these subtrees have to be isomorphic.
// Case 2: The subtrees rooted at these nodes have been "Flipped"
return (recursive(first.left, second.left) && recursive(first.right, second.right)) ||
(recursive(first.left, second.right) && recursive(first.right, second.left));
}This is the code that runs when the iterative method is called to check for isomorphic nodes Because the method couldn't be called within itself, I used stacks to pop nodes and iterate through.
/**
* iteratively checks if the two nodes are isomorphic
*/
static boolean iterative(Node tree1, Node tree2) {
Stack<Node> stack1 = new Stack<>(); //nodes from tree1
Stack<Node> stack2 = new Stack<>(); //nodes from tree2
stack1.push(tree1);
stack2.push(tree2);
Node first, second;
while (!stack1.isEmpty() && !stack2.isEmpty()) {
first = stack1.pop();
second = stack2.pop();
//might be isomorphic continue...
if ((first == null) && (second == null)) continue;
//not isomorphic if comparing null nodes
if ((first == null) || (second == null)) return false;
//if they're not equal they're not isomorphic
if (!(first.data == second.data)) return false;
// There are two possible cases for first and second to be isomorphic
// Case 1: The subtrees rooted at these nodes are "Symmetric".
if (nodesEqual(first.left, second.left) &&
nodesEqual(first.right, second.right)) {
//add the child nodes to the stacks in symmetric order so they'll be pulled symmetrically
stack1.push(first.left);
stack2.push(second.left);
stack1.push(first.right);
stack2.push(second.right);
continue;
}
// Both of these subtrees have to be isomorphic.
// Case 2: The subtrees rooted at these nodes have been "Flipped"
if (nodesEqual(first.left, second.right) &&
nodesEqual(first.right, second.left)) {
//add the child nodes to the stacked in flipped order so they'll be pulled flipped and the nodes match up
stack1.push(first.left);
stack2.push(second.right);
stack1.push(first.right);
stack2.push(second.left);
continue;
}
return false;
}
return true;
}
private static boolean nodesEqual(Node first, Node second) {
// if Both trees are NULL, trees isomorphic by definition
if ((first == null) && (second == null)) {
return true;
} else {
// Exactly one of the first and second is NULL, trees not equal nor are they isomorphic
if ((first == null) || (second == null)) {
return false;
} else {
//if the two are equal trees might be isomorphic
return first.data == second.data;
}
}
}This is the list of classes that this program uses
- This is just a basic class for the nodes that we'll be working with for this project, there isn't much to this class you can take a look at it here [Node]
- This is the implementation of the algorithm to check if nodes are isomorphic
- This is the driver class that has the main method within it. It creates the nodes, prints them and applies both algorithms to 2 different nodes five times
- This class is just a utility class to print out the binary nodes in a visually pleasing format to the console