Question

Let \(F: T \rightarrow S\) be an isomorphism between binary trees \(T\) and \(S\). Let \(v \in T\) be a vertex at level \(k\) for \(k \geq 0\). Prove that \(F(v)\) is at level \(k\) in \(S\).

Short answer

Isomorphism preserves levels, so if \(v\) is level \(k\), then \(F(v)\) is also level \(k\).

Step-by-step solution

Define Isomorphism in Binary Trees

An isomorphism between two binary trees, \(T\) and \(S\), means there exists a bijective function \(F : T \rightarrow S\) that preserves the structure of the trees. Specifically, \(F\) maps vertices of \(T\) to vertices of \(S\) such that if \(x\) is connected to \(y\) in \(T\), then \(F(x)\) is connected to \(F(y)\) in \(S\). Additionally, the order of left and right children is preserved by \(F\).

Concept of Tree Level

In a binary tree, the level of a vertex \(v\) is defined by the number of edges on the path from the root of the tree to the vertex \(v\). The root itself is at level 0. Therefore, if a vertex \(v\) in \(T\) is at level \(k\), there are \(k\) edges from the root of \(T\) to \(v\).

Apply Structure Preservation

Since \(F\) is an isomorphism, it preserves the structure of connections in the trees. This means that there is a direct edge correspondence via \(F\). If \(v\) has a path from the root involving \(k\) edges in \(T\), then \(F(v)\) must also have a path from the root of \(S\) involving the exact same number of edges, as the structure is preserved.

Establishing Level Equality

Since the number of edges from the root of \(T\) to any vertex \(v\) is the same as from the root of \(S\) to \(F(v)\) due to the structure-preserving property of the isomorphism, \(F(v)\) must also be at level \(k\) in \(S\).

Conclusion

We showed that an isomorphism \(F\) preserves not just the connections but the exact depth of vertices from the root. Hence, if a vertex \(v\) in \(T\) is at level \(k\), \(F(v)\) in \(S\) must also be at level \(k\), completing the proof.

Key concepts

Tree Level

In a binary tree, the concept of a level is crucial for understanding the position of vertices. A tree level refers to the number of edges that you need to traverse starting from the root to reach a specific vertex. This gives us a numerical measure of how deep in the tree a vertex is situated.
For instance, the root of a tree is always at level 0 because there are no edges to traverse to reach itself. If a vertex is directly connected to the root (one edge away), it is at level 1. Similarly, a vertex that requires traversing two edges from the root is at level 2, and so forth.
This concept is universally consistent across trees, making it a reliable way to describe the relative position of elements within any binary tree. Knowing the level of each node helps in various computations, such as analyzing tree depth or performing level-order traversals.

Bijective Function

A bijective function is a special type of mapping that is both injective and surjective. Injective, or one-to-one, means that each element of the domain is mapped to a unique element in the codomain. Surjective, or onto, implies that every element of the codomain is the image of at least one element of the domain.
In the context of binary trees, an isomorphism is represented by a bijective function. This establishes a perfect one-to-one correspondence between the vertices of tree \(T\) and tree \(S\). Hence, there are no nodes in tree \(T\) that share an image in tree \(S\), and conversely, every node in tree \(S\) stems from exactly one node in tree \(T\).
The bijection ensures that the trees are essentially the same in structure, though possibly not in content. Each vertex in tree \(T\) matches a distinct vertex in tree \(S\), maintaining both the uniqueness and complete pairing of elements.

Structure Preservation

Structure preservation in binary trees refers to maintaining the overall layout and relationships between nodes during transformations or mappings, like an isomorphism. This ensures that the relative positioning of nodes remains consistent between the two trees.
When an isomorphism maps tree \(T\) onto tree \(S\), every parent-child relationship and sibling order is retained. For example, if a node \(x\) is a parent to \(y\) in tree \(T\), their images \(F(x)\) and \(F(y)\) will hold the same relationship in tree \(S\).
This concept is especially vital in proving properties such as level invariance under the mapping. If \(v\) in \(T\) is at level \(k\), structure preservation guarantees that \(F(v)\) will be placed similarly at level \(k\) in \(S\). Such preservation allows complex operations and proofs to be carried out based on structure rather than content alone.