Prove that if $T$ is a full binary tree with $i$ internal vertices (i.e. non-terminal nodes), then $T$ has $i+1$ terminal vertices (i.e. leaf nodes) and $2i+1$ total vertices.
From the definition of full binary tree, we know that $i_{th}$ node has either left child (i.e. $2i$), or right child (i.e. $2i+1$), and it seems to me that this problem will utilize such property, but I could not exactly relate to it. What are possible approaches for this question?
Update: Here is my solution without handshake lemma, and would be happy yo know whether it is correct or not:
The vertices of $T$ consist of the vertices that are children (of some parent) and the vertices that are not children (of any parent). There is one nonchild—the root. Since there are $i$ internal vertices, each having two children, there are $2i$ children. Thus the total number of vertices of $T$ is $2i + 1$, and the number of terminal vertices is $(2i+1)−i = i+1$.
Note: Just upvote if it is right solution, no need to further comment on this question.
Hint
Let $x$ be number of nodes and $y$ number of edges.
Suppose $x>1$. Root has degree $2$. A leaf has degree $1$. Any non-terminal node that is not root has degree $3$.