Flip Equivalent Binary Trees In C++

In this guide, we will explore the process of reversing equal binary trees in C++ along with its practical application.

By interchanging the positions of the left and right descendants in certain nodes, it is possible to transform two binary trees into each other. This forms the foundation for the idea of equivalent binary trees through flipping.

The trees are considered to be flip equivalent since the swapping of left and right children of the root node can transform the first tree into the second tree.

Operations:

Swapping the left and right children of each node recursively is essential to invert a binary tree.

The existing nodes are the same.

In this scenario, we iteratively verify whether their left branches are flip equivalent and whether their right branches are flip equivalent.

The existing nodes are not identical.

In this scenario, we iteratively verify whether exchanging the left and right subtrees of one tree results in flip equivalent trees with the other tree.

Symmetry

There is symmetry in flip equivalence. Tree-2 is considered flipped equivalent to Tree-1 when Tree-1 and Tree-2 exhibit similarity.

Properties:

This characteristic is a result of the fact that when a node is flipped in a binary tree, it simply exchanges its left and right children. Consequently, even though the tree's arrangement is altered, the positions of nodes at the same level stay the same.

Applications

In scenarios where the sequence of elements holds more significance than their specific values, flip equivalency plays a crucial role. This concept is commonly applied in algorithms to determine outcomes or characterize symmetrical patterns.

It can be utilized in scenarios involving the transformation and comparison of binary trees, especially when the organization of nodes holds significance.

Example:

Let's consider a scenario to demonstrate the concept of flipping binary trees in C++.

#include <iostream>
// A definition of a binary tree node.
struct Treenode
{
    int val;
    Treenode* left;
    Treenode* right;
    Treenode(int x) : val(x), left(nullptr), right(nullptr) {}
};
class Demo
{
public:
    bool Trees_Flip_Equivalent(Treenode* root_1, Treenode* root_2) 
    {
        // Base case: if both roots are null, they are flip equivalent
        if (root_1 == nullptr && root_2 == nullptr)
            return true;
        // They are not flipped equal if one of the roots is null or their values do not match.
        if (root_1 == nullptr || root_2 == nullptr || root_1->val != root_2->val)
            return false;
        // Check flip equivalence with children swapped and not swapped
        return (Trees_Flip_Equivalent(root_1->left, root_2->left) && 
                Trees_Flip_Equivalent(root_1->right, root_2->right)) ||
               (Trees_Flip_Equivalent(root_1->left, root_2->right) && 
               Trees_Flip_Equivalent(root_1->right, root_2->left));
    }
};
int main()
{
    Demo d;
    Treenode* root_1 = new Treenode(7);
    root_1->left = new Treenode(9);
    root_1->right = new Treenode(6);
    root_1->left->left = new Treenode(10);
    root_1->left->right = new Treenode(12);
    Treenode* root_2 = new Treenode(7);
    root_2->left = new Treenode(6);
    root_2->right = new Treenode(9);
    root_2->right->left = new Treenode(12);
    root_2->right->right = new Treenode(10);
    if (d.Trees_Flip_Equivalent(root_1, root_2))
        std::cout << "The trees are flip equivalent." << std::endl;
    else
        std::cout << "The trees are not flip equivalent." << std::endl;
    return 0;
}

Output:

The trees are flip equivalent.

Explanation:

If two binary trees exhibit flip equivalence, this can be verified by employing the TreesFlipEquivalent function within the Demo class in C++ code. The TreesFlipEquivalent function undertakes a recursive comparison of the trees to ascertain flip equivalence, accounting for scenarios where children are exchanged or remain unaltered. Within the main function, two binary trees, root1 and root2, are instantiated even though their structures differ. Upon determining if they are flip equivalent, the relevant process is executed, and the outcome is presented accordingly. In the context of this scenario, the trees are deemed flip equivalent because it is feasible to generate the second tree from the initial tree's root node by interchanging the left and right children.

Complexity Analysis:

Time Complexity

• ✅ The number of nodes and their structures in the binary trees determine the function's time complexity.
• ✅ In the worst case, the code checks the flip equivalency of both trees by repeating through each node.
• ✅ The time complexity is O(n) , where n is the number of nodes in both trees because each node is visited once.

Space Complexity

• ✅ During the function's execution, recursive calls are mostly responsible for the space complexity.
• ✅ The function recurses to the depth of the trees in the worst case.
• ✅ The space complexity of the function is O(h) , where h is the tree height.
• ✅ In the worst-case scenario, the space complexity might be O(n) , where n is the number of nodes in the larger tree if the trees are skewed and out of balance.

The function allocates a fixed amount of additional memory for variables and handles the overhead of function calls, with the exception of recursive calls, which can be disregarded during asymptotic analysis.