Use zod schema defaults when extracting channels

[hntrl]

Privileged issue

• I am a LangGraph.js maintainer, or was asked directly by a LangGraph.js maintainer to create an issue here.

Issue Content

Currently, in order to set a default for a state key when using zod schemas, I have to do something like this:

import * as z from "zod";
import { registry } from "@langchain/langgraph/zod";

const schema = z.object({
  foo: z.string().register(registry, {
    default: () => "bar"
  })
});

const graph = new StateGraph(schema);

But I should be able to set defaults inside of a zod schema using the native method:

import * as z from "zod";
import { StateGraph } from "@langchain/langgraph";

const schema = z.object({
  foo: z.string().default("bar")
});

const graph = new StateGraph(schema);