Bishop_1
Bishop_1

Reputation: 71

Is BFS and a Tree Data Structure Sufficient for Comparing if two Trees are Structurally Equal?

I’m working on a problem where I need to compare multiple lineages (family trees) to check if they are structurally identical. Each lineage starts from a single root (ancestor) and extends downwards. The trees are ordered by the age of the children, and each node has a gender indicator (I, M, K for intersex, male, female, respectively).

The trees are considered structurally equal if:

  1. The root nodes of both trees have the same gender.
  2. The number of children at each node matches between the trees.
  3. The children at each level are ordered the same way, and the nth child of one root is structurally identical to the nth child of the other root.

Here's an example that shows when two trees are not structurally equal.

The problem requires an algorithm with a time complexity of O(n * m), where n is the number of lineages, and m is the number of nodes in the largest tree. We're given that a parent can't have more than 12 children. We're required to use decomposition in our algorithm.

I’ve considered using BFS for tree traversal, as it processes nodes level by level, which fits well with comparing ordered children. I would also use a tree data structure to represent each lineage, where each node contains the gender and references to its children.

However, I’m not entirely sure if this approach is sufficient to meet the problem's requirements, especially given the constraints around ordering and early termination if the structures are not identical.

So to my question: Would using BFS combined with a tree data structure be sufficient to compare these trees in terms of both time complexity and structure? How does BFS start without a common root? Wouldn't that imply a common ancestor and be incorrect for this type of comparison?

Upvotes: 1

Views: 77

Answers (1)

trincot
trincot

Reputation: 350821

You can use a depth-first algorithm, with the exception that you do check rule 2 (the number of children must be equal) before traversing deeper into the tree. That aspect could be considered BFS-like, but that's where the resemblance ends.

The DFS traversal would happen in tandem, i.e. where you move the same way in the two lineages at the same time.

A DFS traversal lends itself for a recursive implementation, whereby you also fulfill the requirement to use "decomposition": the problem for one node is split into the subproblems that the node's children represent.

As suggested, the data structure would consist of nodes that have two properties: a gender and a list of children, where each child is again such a node.

Here is an implementation in JavaScript:

class Person {
    constructor(gender, children=[]) { // Second argument defaults to empty list
        this.gender = gender;
        this.children = children;
    }
}

function areIdentical(lineageA, lineageB) {
    // The root nodes of both trees must have the same gender:
    if (lineageA.gender != lineageB.gender) return false;
    // The number of children at each node matches between the trees
    if (lineageA.children.length != lineageB.children.length) return false;
    for (let i = 0; i < lineageA.children.length; i++) {
        // The nth child of one root is structurally identical to the nth child of the other root
        if (!areIdentical(lineageA.children[i], lineageB.children[i])) return false;
    }
    // All conditions are satisfied
    return true;
}

// Create the lineages as in your example:
const lineageA = new Person("K", [
    new Person("K", [
        new Person("I"),
        new Person("M"),
    ]),
    new Person("M"),
    new Person("K", [
        new Person("K"),
    ]),
]);

const lineageB = new Person("K", [
    new Person("M"),
    new Person("K", [
        new Person("I"),
        new Person("M"),
    ]),
    new Person("K", [
        new Person("K"),
    ]),
]);

const result = areIdentical(lineageA, lineageB);
console.log(result); // false

Upvotes: 1

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