A knot has genus zero exactly when it is the unknot

If $K$ is ambient isotopic to the standard unknot, carry its planar spanning disk through the isotopy. This is a genus-zero spanning surface, and nonnegativity gives $g(K)=0$.

Conversely, use [F1] to let a smooth embedding $e:D^2\to S^3$ parametrize a genus-zero spanning surface with oriented boundary $K$. For $0<r\le 1$ the circles $x\mapsto e(rx)$, $x\in S^1$, are embedded and form a smooth isotopy from $K$ to a sufficiently small circle about $e(0)$. This contracts circles inside the embedded disk; it does not contract the final circle to a point. Choose a point outside this spanning disk and the fixed standard disk; their union has empty interior, since finitely many surface slice charts cover each compact disk. Its stereographic complement is one common positive chart ${\mathbb{R}}^3$ containing both disks. Write $p$ for the chart image of $e(0)$ and $E$ for the expression of $e$ translated by $-p$, so $E(0)=0$. For small fixed $r>0$, the maps $x\mapsto E(trx)\mathrm{/}t$ for $0<t\le 1$, continued at $t=0$ by $rd{E}_0(x)$, are an isotopy of embedded circles in that chart. Indeed Taylor's formula gives the smooth extension and, on a small disk, the derivatives are uniformly close to the injective linear map $rd{E}_0$; projection onto its image plane gives a uniform injectivity estimate by integration along line segments. Their images lie in the chart after decreasing $r$.

The final linear ellipse is isotopic to an oriented round circle in its plane: use the positive square root of its two-dimensional Gram matrix to deform its ordered independent spanning vectors to an orthonormal pair. Complete that pair by its positive normal to a positive ambient frame. Apply the same shrinking and linearization to the fixed standard disk in the common stereographic chart, obtaining an oriented round circle about its interior point $q$. Join the two round circles explicitly in this ${\mathbb{R}}^3$: move the center along the segment from $p$ to $q$, interpolate their positive radii, and rotate the first positive frame to the second through rotations in $SO(3)$ (successive rotations about coordinate axes suffice). Each intermediate circle is embedded, and the whole finite track is compact in the chart. A half-turn about a diameter reverses a circle orientation if required while preserving the ambient orientation. Reverse the standard disk shrinking to finish at the fixed standard unknot. Extend the concatenated circle isotopies by [F2]; each ambient diffeomorphism preserves orientation since it starts at the identity. Thus $K$ is the standard oriented unknot.