This slideshow shows how to reduce an input instance to HAMILTONIAN CYCLE to an equivalent instance of TSP in polynomial time

For a given graph $G = ( V , E )$ , the HAMILTONIAN CYCLE problem is to find whether $G$ contains a Hamiltonian Cycle that is, a cycle that passes through all the vertices of the graph exactly once.

For a given weighted graph $G' = ( V' , E' )$, with non-negative weights, and integer $k'$, the TRAVELING SALESMAN problem is to find whether $G'$ contains a simple cycle of length $\leq k$ that passes through all the vertices. [The length of a cycle is the sum of weights of all the edges in the cycle.]

To reduce HAMILTONIAN CYCLE to TSP for a given graph $G = (V , E)$,
first complete $G$ by adding edges between all pairs of vertices that were not connected in $G$.

Let the new graph be $G'=(V',E')$ where $V' = V$ and $E' = \{(u,v)\}$ for any $u, v \in V'$.

For edges in $G'$ that were also present in $G$ , we assign a weight of $0$.
For the new edges we assign weight $1$

That is , $\forall e=(u,v) \in E'$,
        $ W(e) = 0$, if $(u,v) \in E$.
        $W(e) = 1$, if $(u,v) \not\in E$.

Construction of the complimentary graph can be done in polynomial time.

Let graph $G$ be an input to HAMILTONIAN CYCLE.

The constructed graph $G'$ is as below.
The blue edges were not present in $G$ and so have weight of 1.

The graph $G$ has a Hamiltonian Cycle if and only if there exists a cycle in $G'$ passing through all vertices exactly once, and that has a length $= 0$ (i.e., has a solution for the instance of TRAVELING SALESMAN where $k = 0$.

1. If there is a cycle that passes through all vertice exactly once, and has length $= 0$ in $G'$, the cycle contains only edges that were originally present in graph $G$. (The new edges in $G'$ have weight $1$ and hence cannot be part of a cycle of length $= 0$.)
Hence there exists a Hamiltonian cycle in $G$.

2. If there exists a Hamiltonian Cycle in $G$, it forms a cycle in $G'$ with length $= 0$, since the weights of all the edges is $0$.
Hence there exists a solution for TRAVELING SALESMAN in $G'$ with length $= 0$.

$G'$ has a cycle passing through all vertices exactly once with length $= 0$.

This is a Hamiltonian cycle in G.