您可以使用需要提供JsonConverter的DateTime属性来完成此操作。

public class Root
{
    public int id { get; set; }
    public string name { get; set; }
    public DateTime date { get; set; }
}

var ls = JsonConvert.DeserializeObject<List<Root>>(content, new IsoDateTimeConverter { DateTimeFormat = "MM/dd/yyyy HH:mm:ss" });

反序列化对象声明：

/// <summary>
/// Deserializes the JSON to the specified .NET type using a collection of <see cref="JsonConverter"/>.
/// </summary>
/// <typeparam name="T">The type of the object to deserialize to.</typeparam>
/// <param name="value">The JSON to deserialize.</param>
/// <param name="converters">Converters to use while deserializing.</param>
/// <returns>The deserialized object from the JSON string.</returns>
[DebuggerStepThrough]
public static T? DeserializeObject<T>(string value, params JsonConverter[] converters)
{
    return (T?)DeserializeObject(value, typeof(T), converters);
}

参考：https://www.newtonsoft.com/json/help/html/P_Newtonsoft_Json_Converters_IsoDateTimeConverter_DateTimeFormat.htm

这对我有用

var jsonSettings = new JsonSerializerSettings { DateFormatString = "dd/MM/yyyy hh:mm:ss" };
    
List<Data> data = JsonConvert.DeserializeObject<List<Data>>(json, jsonSettings);

public class Data
{
    public int id { get; set; }
    public string name { get; set; }
    public DateTime date { get; set; }
}